2.3: Basic Concepts
When performing a general survey assessment, nurses use all of their senses to carefully observe the patient. They look at a patient and ask themselves, what are they seeing? They listen to a patient and ask themselves, what are they hearing, both verbally and nonverbally? They smell a patient’s odors and ask themselves, is there anything unusual we need to further assess? They observe a patient’s behaviors and make notes about their functioning and ability to complete daily activities according to their developmental level.
Before performing a general survey assessment, it is important to first ensure the patient is medically stable by completing a brief primary survey. After ensuring the patient is medically stable, a general survey assessment is an overall observation of a patient’s general appearance, behavior, mobility, communication, nutritional, and fluid status. A general survey assessment also includes analyzing height, weight, and vital signs for values that are out of range and require additional follow-up.
Primary Survey
At the beginning of every shift or patient visit, nurses perform a brief primary survey to ensure their patient is medically stable. If any signs indicate patient distress, the provider is notified and emergency care is initiated. For example, changes in level of consciousness or abnormal vital signs often provide early warning signs that a patient’s condition is deteriorating and prompt medical treatment is needed. [1] Mental status, airway, breathing, and circulation are also quickly assessed and emergency actions taken as needed. [2]
In a clinic setting, the patient is observed from the time they are called from the waiting room. Nurses observe a patient’s gait and balance as they walk to the exam room and assess their verbal and nonverbal communication while interacting. If signs of distress are occurring, the nurse follows agency policy and either immediately calls a provider into the room or initiates emergency assistance.
Mental Status
When assessing a patient’s overall mental status, it is important to compare findings to their known baseline if this information is known or available. Initially determine if a patient is responsive or unresponsive. Can you awaken them and are they responding to your questions? Are they oriented to person, place, and time, meaning can they tell you their name, location, and the day of the week? If you are concerned about a sudden change in a patient’s mental status, obtain emergency assistance according to agency policy.
Airway and Breathing
Determine if the patient’s airway is open and if they are breathing adequately. Institute emergency care for respiratory distress as needed. See Figure \(\PageIndex{1}\) [3] for an illustration of checking a patient’s airway, breathing, and circulation (ABCs).
Circulation
If a patient is not responsive, try to awaken them using a sternal rub. A sternal rub is performed by firmly rubbing one’s knuckles on a patient’s sternum to try to elicit a response. If they do not respond, check the carotid pulse and obtain emergency assistance as needed. Briefly observe the color and moisture of their skin. Abnormal findings such as cool, moist, pale, or bluish skin can indicate signs of shock that require emergency care.
General Appearance
After ensuring your patient is medically stable by completing a primary survey, a general survey consists of using your senses to observe a patient’s general appearance, behavior, mobility, and communication. Items to consider when assessing general appearance include the following:
- Signs of pain or distress: Patients may exhibit signs of pain or distress that should be reported to the provider such as grimacing, moaning, or increased anxiety. Set the priorities for your focused assessments based on any signs of distress demonstrated by your patient.
- Age: Observe if the patient appears their stated age. Chronic disease can cause a patient to appear older than their age. Factors can occur with older adults that may influence how well the patient can participate in the assessment, such as hearing, vision, or mobility impairments.
- Body type: No patient is exactly the same; some patients are in good physical shape and others are not. Body type can reflect nutritional status and lifestyle choices.
- Hygiene, grooming, and dress: Observe the overall cleanliness of the patient’s hair, face, and nails and note any odors. Odors can indicate poor hygiene or various disease states. Validate odors by performing additional focused assessments as needed. Note the appearance of the patient’s clothing; is it clean and appropriate for the season? If not, findings can reflect on a patient’s cognitive abilities, emotional state, and ability to complete daily activities.
Behavior
While observing the patient, note their behaviors during your interaction. Consider the following items:
- Affect and mood: People express their mood through facial expressions, eye contact, what they do, and what they say. Eye contact is commonly used to judge a person’s mood because people who are feeling down or depressed commonly avoid eye contact. However, be aware that cultural beliefs may also affect the use of eye contact. Affect refers to the outward display of one’s emotional state. For example, a patient with a “flat affect” refers to very few facial expressions being displayed to indicate emotion, which is often associated with depression. If the patient’s mood or behavior seems inappropriate for their current situation, make a note of that as well. For example, a patient in an usually elated mood in a situation when most people would be seriously concerned can be a symptom of mental illness.
- Family dynamics: If other family members are accompanying the patient, note the characteristics of their interactions. Family dynamics are the patterns of interactions between family members that influence family structure, hierarchy, roles, values, and behaviors. Family dynamics have a strong impact on the way children see themselves, others, and their world. They can also impact the lifestyles of older adults who rely on their children for assistance in activities of daily living and health care.
- Signs of patient abuse: Abuse occurs in many different forms such as physical, emotional, mental, verbal, sexual, economic, or financial. Most states mandate nurses and other health care professionals to report suspected child and elder abuse to the proper authorities. Observe if your patient seems fearful, excessively quiet, or has physical signs of abuse, such as bruising or burn marks. If you suspect abuse, attempt to interview the patient alone. Promptly report your concerns according to agency policy and state mandates.
- Substance use disorder: Approach your patient in a caring and nonjudgmental way, but be aware that unusual signs or behaviors can be indicators of substance use disorder. For example, if a patient’s pupils are unusually dilated or constricted, this can be a sign of substance use disorder. Contact the provider with concerns about substance use disorder.
See Figure \(\PageIndex{2}\) [4] of an image of a man with poor hygiene and nonverbal behavior requiring further assessment, such as disease management, medication management, and ability to complete activities of daily living.
Mobility
Observe your patient’s body movements, noting posture, gait, and range of motion.
- Posture: Patients with normal sitting and standing posture are upright and have a parallel alignment from the shoulders to the hips. Note if the patient is hunched, slumped, contracted, or rigid. See Figure \(\PageIndex{3}\) [5] for an image of a patient with a type of slumped posture called kyphosis.
-
- Gait and balance: Observe how the patient walks or stands. Are the movements organized, coordinated, or uncoordinated? Are they able to maintain balance while standing without leaning on or touching anything? Healthy people walk with a smooth gait and arms moving freely at their sides and are able to stand unassisted. A change in gait or balance often signifies underlying health conditions and increases the risk of falling. See Figure \(\PageIndex{4}\) 6] of a patient learning how to use an assistive device for an altered gait.
Range of motion and mobility: Observe the patient moving their extremities. Do extremities on the right and left sides move equally? Note any tremors or movements that are not purposeful. Are the patient’s abilities appropriate for their age? Is the patient moving normally or do they have specific limitations? Does the patient use any assistive devices such as a cane or walker? Mobility is an important component in being able to care for oneself independently, so note any potential concerns for adults that may impact their ability to complete activities of daily living.
Communication
- Speech: Observe how your patient is speaking during your interaction. Are they speaking in an understandable tone and even pace, or is it garbled or difficult to understand? Neurological disorders can cause speech to be slow, slurred, and hard to understand. Is there an emotional component to their words? Is there a language barrier requiring an interpreter?
- Response to commands: Does the patient follow instructions you are providing during your assessment, or do they have difficulty in understanding or cooperating?
Nutritional Status
Visually observing the patient’s overall nutritional status can provide cues for additional focused assessments related to appetite, diet, food intake, or exercise. Many factors can influence a patient’s nutritional status such as financial or transportation issues, swallowing difficulties, missing teeth, or poorly fitting dentures.
Fluid Status
Observe overall fluid status. Dehydration can be indicated by dry skin, dry mucous membranes, or sunken eyes. Conversely, patients with excess fluid often have swelling or edema in their extremities and may exhibit signs of difficulty breathing.
Height, Weight, and BMI
Height and weight can be used as a guide to reflect the patient’s general health. Weight is routinely assessed during all health care visits. Infants and children are measured to assess their growth and development. Hospitalized patients often have daily weights assessed to monitor for changes in their medical condition. Document findings based on agency policy. For example, in some agencies, height is documented in centimeters and weight is documented in kilograms. Recall that one inch is equivalent to 2.5 centimeters and 1 kilogram is equivalent to 2.2 pounds.
Body Mass Index (BMI) is a standardized reference range that is used to analyze a patient’s weight status and provides a representation of body fat. However, it is important to note that BMI may not be accurate for athletes with increased muscle mass, people with edema or dehydration, or older adults who have lost a significant amount of muscle mass. See a BMI table in Table \(\PageIndex{1}\) . [7] To use the BMI table, find the height in inches in the left column, move across the row to closest weight, and then read the BMI where the column and row intersect. For example, a person who is 5′ 9″ tall is 69 inches. If the patient weighs 155 pounds, the BMI is 23. BMI indicating a healthy weight is between 18.5 to 24.9.
BMI can also be calculated using the formula of BMI = kg/m 2 (weight in kilograms divided by height in meters squared).
The following classifications are used based on a person’s BMI:
Underweight: Below 18.5 kg/m2
Healthy weight: 18.6 to 24.9 kg/m2
Overweight: 25 to 29.9 kg/m2
Obesity: Over 30 kg/m2 to 34.9 kg/m2
Extreme obesity: Over 35 kg/m2
Establish a trusting, nonjudgmental relationship with your patient. Use a calm voice and do not appear rushed. Provide them with your undivided attention, and use all of your senses when interacting to pick up on important cues related to their current health status.
Life Span Considerations During a General Survey Assessment
Children
When performing a general survey on a child, be aware of their developmental stages to establish expectations for normal findings. Use a calm and gentle tone to establish trust. Demonstrations on dolls or stuffed animals before performing procedures are often helpful. Be aware that the parents of toddlers and school-aged children will likely provide most of the information during an assessment.
 It is often helpful to allow time to interview the adolescent privately, in addition to gathering information when the parent is present.
Older Adults
Recognize normal changes associated with aging when performing a general survey. If your patient wears glasses or hearing aids, be sure they are in place before asking questions. You may need to allow for extra time for your assessment, depending on the abilities of your patient. If an older adult is unable to communicate effectively, nurses may also consult a variety of sources such as family members and electronic medical records for more information.
Cultural Adaptations During a General Survey Assessment
Adapt your communication during a general survey assessment to your patient’s cultural beliefs and values. For example, some individuals believe that direct eye contact with authorities is considered disrespectful and avoid eye contact. Other individuals may nod to indicate they are listening but this does not mean they are in agreement with what you are saying. Some patients prefer that a same sex individual perform care.
- This work is a derivative of StatPearls by Toney-Butler and Unison-Pace licensed under CC BY 4.0 ↵
- Chemical Hazards Emergency Medical Management. (2020, April 17). Primary and secondary survey. National Institutes of Health. https://chemm.nlm.nih.gov/appendix8.htm ↵
- “ Checking respiratory.png ” by User:Rama licensed under CC BY-SA 3.0 FR ↵
- “ Homeless man in Los Angeles(7618018076).jpg ” by Alex Proimos licensed under CC BY 2.0 ↵
- “ Posturalkyphosis ” by Lab Science Career licensed under CC BY-NC-SA 2.0↵
- “ US_Navy_071015-N-5086M-202_Retired_Marine_Corps_Cpl._Timothy_Jeffers_walks_on_his_prosthetic_legs_while_using_the_hands-free_harness_walking_gait_training_device_during_a_therapy_session_in_the_new_Comprehensive_Combat_and_Com.jpg ” by U.S. Navy photo by Mass Communication Specialist 2nd Class Greg Mitchell is in the Public Domain ↵
- This work is a derivative of U.S. National Institutes of Health’s National Heart, Lung, and Blood Institute (NHLBI), https://www.wikidoc.org/index.php/File:BMIReferenceChart.jpg licensed under CC BY-SA 3.0 ↵


9.2: Finite Sets
In the previous section, we used the phrase “finite set" without formally defining it. Let’s be a bit more precise. The following shorthand comes in handy.
Definition 9.17. For each \(n\in \mathbb{N}\) , define \([n]:= \{1,2,\ldots,n\}\) .
For example, \([5]=\{1,2,3,4,5\}\) . Notice that our notation looks just like the notation we used for equivalence classes. However, despite the similar notation, these concepts are unrelated. We will have to rely on context to keep them straight.
The next definition should coincide with your intuition about what it means for a set to be finite.
Definition 9.18. A set \(A\) is finite if \(A=\emptyset\) or \(card(A)=card([n])\) for some \(n\in\mathbb{N}\) . If \(A=\emptyset\) , then we say that \(A\) has cardinality 0 and if \(card(A)=card([n])\) , then we say that \(A\) has cardinality \(n\) .
Let’s prove a few results about finite sets. When proving the following theorems, do not forget to consider the empty set.
Theorem 9.19. If \(A\) is finite and \(card(A)=card(B)\) , then \(B\) is finite.
Theorem 9.20. If \(A\) has cardinality \(n\in\mathbb{N}\cup\{0\}\) and \(x\notin A\) , then \(A\cup\{x\}\) is finite and has cardinality \(n+1\) .
Consider using induction when proving the next theorem.
Theorem 9.21. For every \(n\in\mathbb{N}\) , every subset of \([n]\) is finite.
Theorem 9.20 shows that adding a single element to a finite set increases the cardinality by 1. As you would expect, removing one element from a finite set decreases the cardinality by 1.
Theorem 9.22. If \(A\) has cardinality \(n\in\mathbb{N}\) , then for all \(x\in A\) , \(A\setminus \{x\}\) is finite and has cardinality \(n-1\) .
The next result tells us that the cardinality of a proper subset of a finite set is never the same as the cardinality of the original set. It turns out that this theorem does not hold for infinite sets.
Theorem 9.23. Every subset of a finite set is finite. In particular, if \(A\) is a finite set, then \(card(B)<card(A)\) for all proper subsets \(B\) of \(A\) .
Induction is a sensible approach to proving the next two theorems.
Theorem 9.24. If \(A_1,A_2,\ldots, A_k\) is a finite collection of finite sets, then \(\displaystyle \bigcup_{i=1}^k A_i\) is finite.
The next theorem, called the Pigeonhole Principle , is surprisingly useful. It puts restrictions on when we may have an injective function. The name of the theorem is inspired by the following idea: If \(n\) pigeons wish to roost in a house with \(k\) pigeonholes and \(n>k\) , then it must be the case that at least one hole contains more than one pigeon. Note that 2 is the smallest value of \(n\) that makes sense in the hypothesis below.
Theorem 9.25. If \(n,k\in\mathbb{N}\) and \(f:[n]\to [k]\) with \(n>k\) , then \(f\) is not injective.


Scholarly Articles as Sources
Articles in scholarly journals are valued for several reasons. First, they are usually trustworthy because their publication process includes a peer review that helps insure their accuracy and contribution to their disciplines. In addition, they often contain the first reports of new research, which makes their sections on methodology, data, analysis, and interpretation primary sources. Sometimes they instead consist of literature reviews summaries of multiple research studies done in the past on particular subjects of current interest. That makes those articles very helpful secondary sources.
Peer-Reviewed Sources
The most-respected scholarly journals are peer-reviewed, which means that experts in their field other than the author and editor check out each article before it can be published. It’s their responsibility to help guarantee that new material is presented in the context of what is already known, that the methods the researcher used are the right ones, and that the article contributes to the field.
For those reasons, peer-reviewed articles are more likely to be credible. Peer-reviewed journal articles are the official scholarly record, which means that if it’s an important development in research, it will probably turn up in a journal article eventually.
Here
‘s a longer explanation of the peer review process, which concludes that it is good but not perfect.
Parts of a Scholarly Article
The articles you use for your assignments must also be
relevant
to your research question—not just credible.
Reading a scholarly article usually takes some effort. Here’s
how
to do it.
Finding Scholarly Articles
Most scholarly articles are housed in specialized databases. Libraries (public, school, or company) often provide access to scholarly databases by paying a subscription fee for patrons. For instance, OSU Libraries provide access to hundreds of databases via its
Research Databases List
that are made available free to people affiliated with the University. You can search for a journal title in these databases or view a list of databases by subject. For more information, including how to search databases, see Specialized Databases.
Databases that aren’t subject-specific are called general databases.
Google Scholar
is a free general scholarly database available to all who have access to the Internet, and it provides some scholarly articles. For more information, see our section on using Google Scholar.
Tip: Known Article Searching
What if you have a citation for an article you need and now have to find the actual text of the article? Follow these instructions to
Access to a Known Journal Article
.


1.2: Kinds of Inferences
Let’s look at two different inferences:
Inference i:
The sign says only 3 more miles to the coast, I suppose we’re getting close!
Inference d :
The definition of a scab is a union member who works during a strike, Manny is a union member who is working during a strike, so Manny is a scab.
Notice how even if we accept the premise of inference i, we need not accept the conclusion. Ten gazillion different things could make the sign inaccurate. Maybe the coast moved due to erosion or seismic activity, maybe the sign was stolen from its intended location and moved 10 miles inland, maybe the sign was a practical joke in the first place. Who knows?
For inference d, though, it’s not so open-ended. We have a definition and the claim that an individual meets that definition. If the definition of x is d and a is d , then a is an x . If we disagree with the conclusion, we either have to reject the definition, or the description of Manny. We can’t add new information to change the conclusion. Even if Manny is a good guy, or an alien in disguise, or a really pro-union guy, or supporting three kids and a wife with cancer, he’s still a scab if that is in fact the definition of a scab and if that is in fact a true description of Manny. Harsh, but certainly true if the premises are true.
The point of comparing inferences I and d is to see that there are two fundamentally different sorts of inference. We call an inference inductive if the support the premises provide for the conclusion is less than certain—if the premises don’t guarantee the conclusion. We call an inference deductive if the premises provide conclusive support for the conclusion—if they guarantee the conclusion or make the conclusion certain .
Deductive arguments are mathematical arguments like proofs and the like, logical arguments, arguments from definition, etc. If the premises are true and the argumentative structure is good, then the conclusion must be true.
Inductive arguments are arguments from analogy, arguments from qualified authority, causal inferences, scientific hypothetical reasoning, extrapolations from samples, and so on. Even if the argumentative structure is great, the truth of the premises only even makes the conclusion probably true at best.
There’s a third kind of argument where we select the best explanation from all of the available plausible explanations. We won’t spend time on it, but it’s worth noting its existence. It’s sometimes called “abduction.”
Deduction : arguments where the premises guarantee or necessitate the conclusion
- Mathematical Arguments, Logical Arguments, Arguments from Definition
Induction : arguments where the premises make the conclusion probable .
- Analogies, Authority, Causal Inferences, Scientific Reasoning, Extrapolations, etc.
Inference to the Best Explanation or Abduction : arguments where the best available explanation is chosen as the correct explanation.
Validity
Remember that truth is a property of propositions. That is, only propositions can be true or false. Arguments can never be true or false . It simply doesn’t make any sense to claim that an argument is true or false.
Okay, let’s talk about deductive arguments for a hot minute. Deductive argumentative structures are either valid or invalid. An invalid argument structure is one where the premises don’t guarantee the truth of the conclusion, but they should , given the type of argument involved. For instance, if it’s a mathematical argument, then it’s premises should guarantee its conclusion so it’s deductive. But if its premises don’t in fact guarantee its conclusion, then its an invalid deductive argument.
A valid argument structure is an argument structure where the premises guarantee the conclusion. That is, if the premises are true, the conclusion follows necessarily . It’s impossible for the premises to be true and the conclusion false. If 2+2=3, and 6-3=3, then necessarily , beyond any doubt , 2+2=6-3. No ifs, ands, or buts about it. It’s impossible for that conclusion to be false without at least one of those premises being false as well. All true premises and a valid argument means the conclusion must be true.
Keep in mind that validity is about structures . So the previous paragraph’s arithmetical argument has the structure a+b=c, d+e=c, therefore a+b=d+e. Anything we sub in for the letters, if we create two true premises, will necessitate a true conclusion.
What’s a valid argument that has true premises? That’s called a sound argument. Soundness is about both structure and truth: you have to have a good structure and true premises to be a sound argument. An unsound argument, conversely, is an argument that either is invalid or has at least one false premise.
Truth, Validity, Soundness
What’s Truth ? A proposition makes a statement about the world and the world either is or isn’t the way the proposition describes it to be. One proposition claims that the Gross Domestic Product of the United States of America is approximately $14 Trillion. To find out whether this is true or false, go figure out what the GDP of the US is. Is it approximately $14 Trillion? Another proposition claims that there’s a brown cat on the front porch of your house. Is this true? To find out, just go look at the world: is there in fact a cat on your front porch? Is it a brown cat?
The propositions that make up an argument (the premises and the conclusion) are all either true or false. As with all things in philosophy, there is a lot more to say about the complexities here. Some of the earliest philosophy in the Western philosophical tradition is philosophy of logic or language. Aristotle, for instance, asked whether it’s true or false that there will be a sea battle tomorrow. Isn’t it contingent? Aren’t there lots of indeterminant factors involved in determining whether or not a sea battle will in fact take place? If so, it seems like that proposition is neither true nor false yet. So not every proposition is either true or false. We need not, though, deal with such issues. We can proceed as if every proposition is determinately either true or false.
Remember : Validity is a property of argument structure: it means “this structure is such that if the premises of any argument with this structure are true, then the conclusion of that argument must be true.”
It means: arguments of this structure will never have all true premises and a false conclusion. The structure guarantees the truth of the conclusion given the truth of the premises. Almost like the structure carries the truth of the premises directly to the conclusion without fail. A reliable one-way transporter of truth.
A sound argument is an argument that has a valid structure but then also has true premises. If an argument is sound, and if validity means the conclusion must be true if the premises are true, then the conclusion must be true, then what do we know about the truth of the conclusion of any sound argument? Yes! You’re so smart: the conclusion of any sound argument is guaranteed to be true.
Truth : propositions are either true or false
Validity : good deductive argument structure : True premises make the conclusion necessarily true. (if not, it’s an Invalid structure)
Soundness : Valid deductive argument, all True premises. (If not, it’s an Unsound argument)
\[\fbox{All True Premises} + \fbox{Valid Structure} = \fbox{Sound Argument}\nonumber\]
Truth, Strength, Cogency
Switching over to inductive arguments, we find an analogous set of properties. Again, inductive arguments are made up of propositions, which can be true or false.
The biggest difference is that even good inductive arguments only offer probabilistic support for their conclusions. Meaning accepting all of the premises doesn’t necessitate that one accept the conclusion, it merely gives one more or less strong reason for accepting the conclusion. So the argumentative structure of an inductive argument isn’t either good or bad, it’s a matter of degree (and often a matter of what the actual content is). An inductive argument can therefore offer stronger or weaker inductive support for its conclusion.
“Cogent” and “uncogent” are the words we use in place of “sound” and “unsound” for inductive arguments since inductive arguments cannot be sound or unsound. Cogent, therefore, means all true premises and the premises give strong inductive support for the conclusion.
Consider these two arguments:
|
I saw a black cat Therefore all cats are black |
I saw the Sun rise in the East every day of my life and everyone I know reports the same and history books and ancient astronomers report the same, so the Sun will rise in the East tomorrow. |
Notice how the argument on the left provides pretty weak support for the conclusion. I can believe that the speaker in fact saw a black cat and still think that’s a bad reason for concluding that all cats are black. The right argument, though, is much stronger. There’s much more evidence and the nature of the evidence makes the conclusion much more probable given the truth of all of the premises.
Truth : propositions are either true or false
Strength : Inductive argument : true premises make conclusion probably true.
Cogency : Strong inductive argument, all True premises.
\[\fbox{All True Premises} + \fbox{Strong Inductive Support} = \fbox{Cogent Argument}\nonumber\]
Again, inductive arguments are collections of propositions (the premises and the conclusion(s) are all propositions. And each of these propositions might be either true or false depending on whether it accurately describes reality.
An inductive argument cannot be valid. Why? Because a valid argument guarantees the truth of the conclusion. But an inductive argument only justifies its conclusion to some level of probability .


37.2: Sample Lab Report- Sugar Size and Diffusion Through a Mock-Cell Membrane
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- Last updated
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- Darcy Ernst, May Chen, Katie Foltz, and Bridget Greuel
- Evergreeen Valley College via Open Educational Resource Initiative at Evergreen Valley College
Introduction
Diffusion is the process in which a substance moves from an area of high concentration to an area of lower concentration. It is important for membranes to be semi-permeable. If membranes were universally permeable, they would not legitimately serve their purpose as membranes; certain substances need to be kept out of a cell, and others kept in. If membranes were not at all permeable, there would be no interface between the cell and its environment – effectively starving the cell. Membranes, being selectively permeable, allow in nutrients and other necessary substances, and also provide for the purging of cell waste.
This experiment investigates the permeability of cell membranes to various types of sugars: polysaccharides, disaccharides, and polysaccharides. Dialysis tubing is used to simulate a cell membrane; it is permeable to small molecules and water, but not to larger molecules.
Given the generally larger size of polysaccharides, it is hypothesized that starch will not pass through the dialysis tubing, and that iodine will pass through the membrane due to the small size of its molecules. Based on the trouble that some people have digesting lactose, it is predicted that it is a polysaccharide or disaccharide and will yield diffusion results similar to starch.
Methods
I) Permeability of cell model to starch
The 100ml graduated cylinder was used to measure out 50ml of tap water; the water was poured into a 250ml beaker. One teaspoon of corn starch was added to the beaker and stirred with a spoon.  A piece of dialysis tubing was placed into the beaker of water until it became soft and pliable. The tubing was then extracted from the beaker; one end was tied closed with dental floss, using a double-knot. The other end of the dialysis tubing was opened; a pipette was used to fill the dialysis tubing with the starch solution. Another piece of dental floss was used to tie the end of the dialysis tubing closed.  Another pipette was used to add 15 drops of iodine to the beaker; the solution was mixed with a spoon. The filled dialysis tube was placed into the 250ml beaker so that the cornstarch mixture was submerged in the iodine water mixture. After 15 minutes had passed, results were observed and recorded.
II) Permeability of cell model to lactose
Part I: Determining the type of sugar being tested
The 100ml graduated cylinder was used to measure out 100ml of tap water; the water was poured into a 250ml beaker. Two teaspoons of the lactose was added to the water, and the solution was stirred with a spoon for thorough mixing. 50ml of the resultant solution was measured out using the 100ml graduated cylinder, and was reserved for Part II.
Twenty drops of Benedict’s reagent were placed in a clean, empty test tube. Twenty drops of the lactose solution were added to the same test tube, and the solution was heated in a boiling water bath for 2 minutes. The results were then interpreted.
Twenty drops of Barfoed’s reagent were placed in a clean, empty test tube. Twenty drops of lactose solution were added to the same test tube. The resultant solution was headed in a boiling water bath for 3.5 minutes. The results were then interpreted.
Part II: Permeability of cell model
This method is based on the premise of the unknown sugar being a disaccharide.
 A piece of dialysis tubing was placed in the beaker of water and left to soak until it became soft and pliable. The dialysis tubing was then removed from the beaker, and one end tied closed with a double-knotted piece of dental floss. The other end of the dialysis tubing was opened. The tubing was filled with lactose solution (set aside from Part I); a pipette was used to transfer the solution from the graduated cylinder to the tubing. A second piece of dental floss was used to tie the other end of the dialysis tubing closed. A second 250ml beaker was filled halfway full with tap water.
15 drops of iodine were added to the tap water in the beaker. The resulting solution was swirled with a spoon to mix it; the colors of the baggie solution and the beaker solution were noted. The dialysis tubing baggie was placed in the 250ml beaker so that the lactose solution was submerged in the beaker solution, and left to sit undisturbed for 15 minutes. The color of the baggie solution was noted.
The baggie was removed from the beaker and samples of the beaker solution were transferred to separate, appropriately marked test tubes. 20 drops of Bendict’s reagent were added to one test tube, and the tube heated for 2 minutes. The color of the resulting solution was noted. 20 drops of Barfoed’s reagent were added to the second test tube, and the tube heated for 3.5 minutes. The color of the resulting solution was noted.
Results
| Table 1: Starch experiment results | ||
|---|---|---|
| Solution in baggie | Solution in Beaker | |
| Starting color | Murky white | Clear yellow |
| Color after 15 minutes | Dark purple | yellow |
In the starch experiment as seen in Table 1, the starch solution inside of the dialysis baggie was initially a murky white color. The solution in the beaker, external to the baggie was a clear yellow color. After 15 minutes of submersion in the beaker solution, the baggie had turned a dark purple color. The beaker solution remained clear and yellow.
In Part I of the lactose experiment, the lactose solution was initially a dark brown color. Benedict’s reagent is pale blue in color. Lactose, mixed and heated with the Benedict’s reagent, yielded a solution of a murky yellow-brown color. Barfoed’s reagent, like Benedict’s reagent, is pale blue in color. Lactose, mixed and heated with the Barfoed’s reagent, yielded a pale blue solution.
| Table 2: Lactose experiment results | ||
|---|---|---|
| Solution in baggie | Solution in Beaker | |
| Starting color | brown | yellow |
| Color after 15 minutes | yellow | (? Not given) |
In Part II of the lactose experiment, as seen in Table 2, the lactose solution inside of the dialysis baggie was initially dark brown in coloration. The iodine and water solution in the beaker was a clear yellow color. A Benedict’s test on the beaker solution after the experiment yielded a dark brown liquid; a Barfoed’s test on the beaker solution after the experiment resulted in a clear blue liquid.
Discussion
Permeability of cell model to starch
The starch solution inside of the dialysis baggie went from a murky white color to dark purple; iodine from the beaker solution must have diffused into the dialysis baggie, reacting with the starch solution and producing the “positive” dark-purple result, confirming the presence of a polysaccharide inside of the baggie. The beaker solution remained a clear yellow color throughout the experiment; it can hence be inferred that no polysaccharide was present in the beaker solution at the end of the experiment, and in turn, that no starch diffused out of the baggie and into the beaker solution during the 15-minute soaking.
The experimental hypothesis for this section was correct; starch was unable to diffuse through the cell model, however, iodine was able to diffuse through the cell model. The discrepancy in permeability is due to the difference in the sizes of iodine and starch molecules.
Permeability of cell model to lactose
The Benedict’s test control on lactose yielded a solution that was a murky yellow-brown color; this indicated the presence of a mono- or di- saccharide. The Barfoed’s test control on lactose yielded a solution that was pale blue in color, without any red precipitate; this indicates that no monosaccharaides were present, and in turn, that lactose is a disaccharide. The solution inside of the dialysis tubing changed color in the course of the experiment; this implies that iodine diffused into the dialysis tubing and reacted with the lactose solution. The resulting clear yellow color indicates that there were no polysaccharides present inside of the dialysis tubing.
A negative Benedict’s test is of blue coloration; a test on the beaker solution after the experiment is a very dark red-orange-brown color that looks similar to the original lactose in the tubing. A Barfoed’s test on the post-experiment beaker solution was a clear light blue; no monosaccharaides diffused out into the beaker solution, but this result was irrelevant. The Benedict’s test revealed that lactose was able to diffuse out of the dialysis baggie, into the beaker solution. If the cell model is reliable, it appears that lactose is able to diffuse in and out of cells.
The experimental hypothesis for this section appears to have been wrong; the cell model was permeable to lactose.
Overall, the cell model has demonstrated impermeability to large molecules such as polysaccharides, and permeability to smaller molecules such as disaccharides and iodine molecules. Since the model was permeable to a disaccharide, it would be reasonable to infer that the model will be permeable to monosaccharaides, as they are even smaller in size than disaccharides. Further testing with a variety of disaccharides should be done, to determine whether lactose is unique or whether the cell model is permeable to all disaccharides.
LICENSES AND ATTRIBUTIONS
CC LICENSED CONTENT, ORIGINAL
- Biology 102 Labs. Authored by : Lynette Hauser. Provided by : Tidewater Community College. Located at : http://www.tcc.edu/ . License : CC BY: Attribution


11.6: Turkic Migrations
Nomadic migration was the first major external influence that would be integrated into the culture of the region, as steppe peoples imparted a lasting impression to Central Asia. Beginning with the Xiongnu (209 BCE – 93 CE), a long-term exodus of steppe peoples spread out of Mongolia and into Central Asia. For millennia prior to the rise of Genghis Khan, the winners of the tribal battles for predominance on the Orkhon Steppe , prime pastureland located in western Mongolia, forced the vanquished off to the west. These periodic mass departures of Turkic tribes out of the area progressed southwest into Central Asia in a migration conquest, not a trade diaspora, as happened to Africans during the slave trade. These new arrivals forever altered the ethnic makeup of Central Asia. Previously, the region had been predominantly Persian and Indo-European; when the waves of Turkic tribes penetrated into the area, though, they occupied the great steppe and agricultural basin of Central Asia and pushed these Persian groups to the fringes. Over time, they slowly Turkified the area, endowing it with a more nomadic character.
These Turkic tribesmen divided their society into five strata. Members of the royal tribal clan presided over the social order. This dominant group bestowed its name on the tribal confederation, a collection of tribes. Positioned below them were their allies and associated tribes. Next were the common herders who did not participate in struggles for power. Lower still were the artisans, such as blacksmiths and leatherworkers. And finally, we find slaves at the bottom of the hierarchy. They usually acquired their lowly position in society by means of capture in times of war.
These Turkic wanderers belonged to an unstable confederation of clans and tribes roaming the steppe, loosely bound under a khagan , a charismatic monarch who laid claim to some sort of divine providence. Khagan made use of their personal charisma, as well as their political and military smarts, in order to maintain group cohesion and ward off challenges to their authority. Under strong khagans, tribal confederations were capable of wielding incredible power, but, more often than not, they were notoriously volatile and often imploded upon the death of their leader, collapsing into a brutal struggle for power. The winners in this struggle forced the losers out of the area, and while many went to the north or south, most to the west. Victorious tribes remained in Mongolia on the highly-prized Orkhon Steppe, located near Lake Baikal.
Although the khaganate was a diarchy, or system of dual rule, with the oldest son controlling half of the land, it lacked a clear transition of power, like hereditary succession. Because the khagan theoretically ruled over a series of tribal confederations, any member of the tribal confederation could ascend to the position of monarch by demonstrating their personal charisma and martial skills on the battlefield. This often resulted in a fight to prove oneself that could erupt into broader inter-tribal strife.
Periodic Turkic migrations into Central Asia transformed the sedentary culture of the region. These steppe peoples lived by practicing pastoral nomadism, a way of life centered around herding that most likely predated the Turks but was eventually adopted by them. Their culture was utilitarian in nature and provided all the necessities for life on the great plains of Central Asia, including food, clothing, shelter, and transportation. In order to maintain their pastures, these horsemen followed a fixed, seasonal pattern of migration because they did not want their flocks to overgraze. During the winter, for instance, they camped in foothills and mountain valleys, where it was warmer at lower altitudes. There they built fixed shelters with one main objective: survival. The oral tradition, which included songs, epic narratives, and parables, flourished during the inhospitable winter months.
In the spring, the nomads made a ten-day trip to the prairieland to graze their herds on fresh grass that just emerged from mountain runoff. There the women and children erected a central camp, usually comprising four yurts, while the men divided the flocks into their specific pastures. They established about ten satellite camps around the central camp, with each herd positioned about ten to twenty miles from the center. This separation of camps minimized the potential threat that their enemies posed to their herds. During the summer, they traveled to mid-mountain fields, where it was cooler and offered access to water. Covering about ten miles per day, it took them approximately fifty days to reach this campground. Finally, in the fall, they returned to the steppe in order to make provisions for the harsh winter. These preparations included drying and preserving their meat, and taking milk from their animals.
Enhanced mobility was the key to the survival of pastoral nomads. They actually spent a good portion of their lives on horseback and were accustomed to moving over long distances, taking all of life’s necessities with them. This allowed them to retreat quickly from rival attacks or areas afflicted by natural disaster. Though their way of life appeared seemingly innocuous, it enhanced the ability of these horsemen to expand rapidly and conquer neighboring groups. It was in this manner that pastoral nomadism accorded its practitioners certain martial advantages. The annual Great Hunt served as a military proving ground that helped them hone their fighting skills. In preparation for winter, tribes deployed groups of mounted men, who dispersed in different directions, with the intent of driving every animal within a set perimeter inwards to converge at a pre-established central point. With great coordination taking place over vast distances, these migrants learned how to coordinate their movements based on a color scheme of arrows and whistling patterns. Their herding tactics easily translated to military tactics and proved devastating in combat.
Nomad society was certainly capable of waging war. Their ability to shoot from horseback provided them with a mobile and lethal means to overcome slower, infantry-based armies. These horsemen carried portable, three-foot-long recurve bows capable of piercing enemy armor from over 450 meters. Metal thumb rings enabled a rapid rate of fire without damaging the archer’s fingers. Raised hunting and herding from horseback, nomads even learned how to sleep in the saddle of the Mongol Horse , their indigenous horse. Though not tall in stature, these sturdy mounts displayed impressive endurance and allowed groups to traverse great distances, often up to 160 kilometers per day. The speed with which they could cover territory on their steeds often confused sedentary forces and multiplied the terror factor. Native to the region, these horses were able to forage for themselves and survive on their own. Nomads did not require supply lines and could, therefore, remain on campaign for an average of three years. The combination of the skills acquired from herding, the double-compound bow, and the Mongol Horse, translated to a formula for political domination of Central Asia, at least until the arrival of Genghis Khan and the Mongols.
Turkic domination of the region began on the battlefield, where the strategies of steppe warfare proved devastating to infantry-based armies. The first stage of the nomad battle strategy often commenced with a feigned retreat, in which a group of their cavalry engaged the adversary, retreated, and encouraged their opponents to follow them. This technique lengthened the lines of their challengers, as they pursued the “retreating” Turkic cavalrymen, who were busy shooting backwards from horseback. The next stage of battle involved outflanking the enemy and enveloping them. They then showered their foes with arrows, the objective being to pin the opponent in place. This alone was often enough to break a sedentary power. When fighting against another steppe power, their reserves charged the opponent’s lines so as to break their forces into pieces and finish them off piecemeal. Most importantly, because of their limited numbers, the Turkic horsemen were reluctant to risk fighting an enemy that they did not believe they could defeat, instead, they would poison water wells, scorch the earth, and retreat. The Mongols would later employ similar battle tactics that allowed them to conquer the whole of Central Asia.


10.25: Reading- Defining Product
A product is a bundle of attributes (features, functions, benefits, and uses) that a person receives in an exchange. In essence, the term “product” refers to anything offered by a firm to provide customer satisfaction, tangible or intangible. Thus, a product may be an idea (recycling), a physical good (a pair of jeans), a service (banking), or any combination of the three. [1]
Broadly speaking, products fall into one of two categories: consumer products and business products (also called industrial products and B2B products). Consumer products are purchased by the final consumer. Business products are purchased by other industries or firms and can be classified as production goods —i.e., raw materials or component parts used in the production of the final product—or support goods —such as machinery, fixed equipment, software systems, and tools that assist in the production process. [2] Some products, like computers, for instance, may be both consumer products and business products, depending on who purchases and uses them.
The product fills an important role in the marketing mix because it is the core of the exchange. Does the product provide the features, functions, benefits, and uses that the target customer expects and desires? Throughout our discussion of product we will focus on the target customer. Often companies become excited about their capabilities, technologies, and ideas and forget the perspective of the customer. This leads to investments in product enhancements or new products that don’t provide value to the customer—and, as a result, are unsuccessful.
Four Levels of the Product
There are four levels of a product (shown in the figure below): core, tangible, augmented, and promised. Each is important to understand in order to address the customer needs and offer the customer a complete experience.
The Core Product
The core product satisfies the most basic need of the customer. For example, a consumer who purchases a healthy snack bar may be seeking health, convenience, or simply hunger relief. A student who buys low-priced, sturdy sneakers may just be seeking footwear. A student on a tight budget who buys top-of-the-line sneakers might be hoping to achieve status. Or, the student might be seeking a sense of freedom by splurging on an item that represents a true sense of style, even though he can’t really afford it. Footwear, status, and freedom are all legitimate core products. The core product is complex because it is so individualized, and, often, vague. The marketer must have a strong understanding of the target customer (and the different segments of target customers) in order to accurately identify the core product.
The Tangible Product
Once the core product has been identified, the tangible product becomes important. Tangible means “perceptible by touch,” so the tangible aspects of a product are those that can be touched and held. This idea can be expanded to also include the characteristics of the product that directly touch the buyer in the buying decision. These are the product elements that the customer will use to evaluate and make choices: the product features, quality level, brand name, styling, and packaging. Every product contains these components to a greater or lesser extent, and they are what the consumer uses when evaluating alternatives.
The importance of each aspect of the tangible product will vary across products, situations, and individuals. For example, at age twenty, a consumer might choose a particular brand of new car (core product=transportation) based on features such as gas mileage, styling, and price (choice=Toyota Yaris); at age forty-five, the core product remains the same, while the tangible components such as quality level, power, features, and brand prestige become important (choice=Audi A6).
The Augmented Product
Every product is backed up by a host of supporting services. The augmented product includes the tangible product and all of the services that support it. Often, the buyer expects these services and would reject the tangible product if they were not available. For example, if you shop at a department store, you are likely focused on a core and tangible product that centers on the merchandise, but you will still expect the store to have restrooms, escalators, and elevators. Dow Chemical has earned a reputation as a company that will bend over backward in order to service an account. It means that a Dow sales representative will visit a troubled farmer after hours in order to solve a serious problem. This extra service is an integral part of the augmented product and a key to their success.
When the tangible product is a service, there is still an augmented product that includes support services. Westin hotels offer hotel nights with a specific set of features as their tangible product. The augmented product also includes dry cleaning services, concierge services, and shuttle services, among others.
In a world with many strong competitors and few unique products, the augmented product is gaining ground, since it creates additional opportunities to differentiate the product from competitive offerings.
The Promised Product
The outer ring of the product is referred to as the promised product . Every product has an implied promise, which is a characteristic that is attached to the product over time. The promised product is the long-term result that the customer hopes to achieve by selecting the product. The promised product may be financial—the resale value of a car, home, or property, for example—but it is often more aspirational. The customer hopes to be healthier, happier, more productive, more successful, or enjoy greater status.
Like the core product, the promised product is highly personal. Generally, marketers find that there will be groupings of customers seeking a similar promise but that there is not a single promised product across all customers.
Can the core product and the promised product be the same thing? Yes, they can, but often the the core product is more focused on the immediate need and the promised product has a longer-term element.
Let’s compare two different examples of the same purchase to understand how the product levels might change for different customers.
| Impetus to Buy | I need to be in Miami for a meeting next Thursday | I need a break from my stressful life |
|---|---|---|
| Core Product | Transportation | Escape, peace of mind |
| Tangible Product |
Airline ticket from New York to Miami
|
Airline ticket from New York to Miami
|
| Augmented Product |
|
|
| Promised Product | Productivity, convenience, success | Escape, peace of mind, happiness |
In the first case, the customer’s impetus to buy is transportation, so that is the core product. In the second case, the purchase is more aspirational and less concrete, so the core product and the promised product are quite similar.
For a marketer, the most important element is to have a holistic view of the product. If I believe that I’m simply selling airline tickets, then I fail to provide the full product offering that will satisfy either of my customers in the example above. And of course, it is always key to truly understand the motivation and perspective of the target customer.
- www.ama.org/resources/Pages/Dictionary.aspx?dLetter=P#product ↵
- www.businessdictionary.com/definition/industrial-goods.html ↵
Contributors and Attributions
- Revision and adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
- Chapter 7: Introducing and Managing the Product, from Introducing Marketing. Authored by : John Burnett. Project : Global Text. License : CC BY: Attribution
- New Kicks. Authored by : Grace Miller. Located at : https://www.flickr.com/photos/gracemillerphotography/6101851438/ . License : CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives


5.2: Articles
The Promise of Care Coordination: Transforming Health Care Delivery
Families USA, April 2013.
Part of a series to help advocates understand and engage in key health system reform issues.
[embeddoc url=” http://familiesusa.org/sites/default...ordination.pdf ” download=”all”]
Services Offered by Private Health and Patient Advocates
Tricia Torrey, August 2016.
Patient Advocate” is a very broad title that encompasses many types of patient advocacy services. The descriptions in this article will help you understand better the patient advocacy services a patient may need.
What You Need To Know: What Is Aging Life Care?
Aging Life Care Association.
Aging Life Care, also known as geriatric care management, refers to a client-centered approach to caring for older adults or others facing ongoing health challenges. This article highlights some of the services that may be provided by an aging life care professional.
Geriatric Care Management – Questions and Answers
Alzheimer’s Association, 2008.
A geriatric care manager coordinates care and services to meet the full social, emotional, physical, and healthcare needs of an older adult. However, care managers do not specialize in all areas. This article looks at some of the ways a geriatric care manager may help older adults and families.
[embeddoc url=” http://www.alz.org/stl/documents/GCM_Tips.pdf ” download=”all”]


3.10: Exercise Supplement
Exercise Supplement
Signed Numbers
For the following problems, find \(−a\) if \(a\) is
how many exercises?
\(27\)
- Answer
-
\(-27\)
\(-15\)
\(-\dfrac{8}{9}\)
- Answer
-
\(\dfrac{8}{9}\)
\(-(-3)\)
\(k\)
- Answer
-
\(-k\)
Absolute Value
Simplify the following problems.
\(|8|\)
\(|-3|\)
- Answer
-
\(3\)
\(-|16|\)
\(-(-|12|)\)
- Answer
-
\(12\)
\(-|0|\)
Addition of Signed Numbers - Multiplication and Division of Signed Numbers
Simplify the following problems.
\(4 + (-6)\)
- Answer
-
\(-2\)
\(-16 + (-18)\)
\(3 - (-14)\)
- Answer
-
\(17\)
\((-5)(2)\)
\((-6)(-3)\)
- Answer
-
\(18\)
\((-1)(-4)\)
\((4)(-3)\)
- Answer
-
\(-12\)
\(\dfrac{-25}{5}\)
\(\dfrac{-100}{-10}\)
- Answer
-
\(10\)
\(16 - 18 + 5\)
\(\dfrac{(-2)(-4) + 10}{-5}\)
- Answer
-
\(-\dfrac{18}{5}\)
\(\dfrac{-3(-8+4)-12}{4(3+6)-2(-8)}\)
\(\dfrac{-1(-3-2)-4(-4)}{-13+10}\)
- Answer
-
\(-7\)
\(-(2-10)\)
\(0 - 6(-4)(-2)\)
- Answer
-
\(-48\)
Multiplication and Division of Signed Numbers
Find the value of each expression for the following problems.
\(P = R - C\). Find \(P\) if \(R = 3000\) and \(C = 3800\).
\(z = \dfrac{x-u}{s}\). Find \(z\) if \(x = 22, u = 30\), and \(s = 8\).
- Answer
-
\(-1\)
\(P = n(n-1)(n-2)\). Find \(P\) if \(n = -3\).
Negative Exponents
Write the expressions for the following problems using only positive exponents.
\(a^{-1}\)
- Answer
-
\(\dfrac{1}{a}\)
\(c^{-6}\)
\(a^3b^{-2}c^{-5}\)
- Answer
-
\(\dfrac{a^3}{b^2c^5}\)
\((x+5)^{-2}\)
\(x^3y^2(x-3)^{-7}\)
- Answer
-
\(\dfrac{x^3y^2}{(x-3)^7}\)
\(4^{-2}a^{-3}b^{-4}c^5\)
\(2^{-1}x^{-1}\)
- Answer
-
\(\dfrac{1}{2x}\)
\((2x+9)^{-3}7x^4y^{-5}z^{-2}(3x-1)^2(2x+5)^{-1}\)
\((-2)^{-1}\)
- Answer
-
\(\dfrac{1}{-2}\)
\(\dfrac{1}{x^{-4}}\)
\(\dfrac{7x}{y^{-3}z^{-2}}\)
- Answer
-
\(7xy^3z^2\)
\(\dfrac{4c^{-2}}{b^{-6}}\)
\(\dfrac{3^{-2}a^{-5}b^{-9}c^2}{x^2y^{-4}z^{-1}}\)
- Answer
-
\(\dfrac{c^2y^4z}{9a^5b^9x^2}\)
\(\dfrac{(z-6)^{-2}}{(z+6)^{-4}}\)
\(\dfrac{16a^5b^{-2}}{-2a^3b^{-5}}\)
- Answer
-
\(-8a^2b^3\)
\(\dfrac{-44x^3y^{-6}z^{-8}}{-11x^{-2}y^{-7}z^{-8}}\)
\(8^{-2}\)
- Answer
-
\(\dfrac{1}{64}\)
\(9^{-1}\)
\(2^{-5}\)
- Answer
-
\(\dfrac{1}{32}\)
\((x^3)^{-2}\)
\((a^2b)^{-3}\)
- Answer
-
\(\dfrac{1}{a^6b^3}\)
\((x^{-2})^{-4}\)
\((c^{-1})^{-4}\)
- Answer
-
\(c^4\)
\((y^{-1})^{-1}\)
\((x^3y^{-4}z^{-2})^{-6}\)
- Answer
-
\(\dfrac{y^{24}z^{12}}{x^{18}}\)
\((\dfrac{x^{-6}}{y^{-2}})^{-5}\)
\((\dfrac{2b^{-7}c^{-8}d^4}{x^{-2}y^3z})^{-4}\)
- Answer
-
\(\dfrac{b^{28}c^{32}y^{12}z^4}{16d^{16}x^8}\)
Scientific Notation
Write the following problems using scientific not
8739
73567
- Answer
-
\(7.3567 \times 10^4\)
21,000
746,000
- Answer
-
\(7.46 \times 10^5\)
8866846
0.0387
- Answer
-
\(3.87 \times 10^{-2}\)
0.0097
0.376
- Answer
-
\(3.76 \times 10^{-1}\)
0.0000024
0.000000000000537
- Answer
-
\(5.37 \times 10^{-13}\)
46,000,000,000,000,000
Convert the following problems from scientific form to standard form.
\(3.87 \times 10^5\)
- Answer
-
387,000
\(4.145 \times 10^4\)
\(6.009 \times 10^7\)
- Answer
-
60,090,000
\(1.80067 \times 10^6\)
\(3.88 \times 10^{-5}\)
- Answer
-
0.0000388
\(4.116 \times 10^{-2}\)
\(8.002 \times 10^{-12}\)
- Answer
-
0.000000000008002
\(7.36490 \times 10^{-14}\)
\(2.101 \times 10^{15}\)
- Answer
-
2,101,000,000,000,000
\(6.7202 \times 10^{26}\)
\(1 \times 10^6\)
- Answer
-
1,000,000
\(1 \times 10^7\)
\(1 \times 10^9\)
- Answer
-
1,000,000,000
Find the product for the following problems. Write the result in scientific notation.
\((1 \times 10^5)(2 \times 10^3)\)
\((3 \times 10^6)(7 \times 10^7)\)
- Answer
-
\(2.1 \times 10^{14}\)
\((9 \times 10^2)(3 \times 10^{75})\)
- Answer
-
\(2.7 \times 10^{78}\)
\((1 \times 10^4)(1 \times 10^5)\)
\((8 \times 10^{-3})(3 \times 10^{-6})\)
- Answer
-
\(2.4 \times 10^{-8}\)
\(9 \times 10^{-5})(2 \times 10^{-1})\)
\((3 \times 10^{-2})(7 \times 10^2)\)
- Answer
-
\(2.1 \times 10^1\)
\((7.3 \times 10^4)(2.1 \times 10^{-8})\)
\((1.06 \times 10^{-16})(2.815 \times 10^{-12})\)
- Answer
-
\(2.9839 \times 10^{-28}\)
\((9.3806 \times 10^{52})(1.009 \times 10^{-31})\)


8.5: Amortized Loans
In Section 8.2, we learned how to find the future value of a lump sum, and in Section 8.4, we learned how to find the future value of an annuity. With these two concepts in hand, we will now learn how to amortize a loan , and how to find the present value of an annuity .
Amortize a Loan
If a person or business needs to buy or pay for something now (a car, a home, college tuition, equipment for a business) but does not have the money, they can borrow the money as a loan.
They receive the loan amount now, called the principal, P , (or present value), and are obligated to pay back the principal in the future over a stated amount of time (term of the loan), as regular periodic payments, PMT , plus interest.
Find the monthly payment for a car costing $15,000 if the loan is amortized over five years at an interest rate of 9%.
Solution
Consider the following scenario:
Two people, Mr. Cash and Mr. Credit, go to buy the same car that costs $15,000. Mr. Cash pays cash and drives away, but Mr. Credit wants to make monthly payments for five years.
Our job is to determine the amount that Mr. Credit needs to pay each month for 5 years. We reason as follows:
If Mr. Credit pays PMT dollars per month, then the PMT dollar payment deposited each month at 9% for 5 years should yield the same amount as the $15,000 lump sum deposited in an annuity for 5 years.
Again, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like them to be the same.
Since Mr. Cash is paying a lump sum of $15,000, the future value \(F\) is given by the lump sum formula (8.2), and it is
\[F =\ 15,000\left(1+\dfrac{.09}{12}\right)^{60} \nonumber\]
Mr. Credit wishes to make a sequence of payments of \(PMT\) dollars per month, and the future value is given by the annuity formula (8.4), and this value is
\[F = PMT\frac{\left[\left(1+\dfrac{.09}{12}\right)^{60}-1\right]}{\dfrac{.09}{12}} \nonumber\]
We set the two future amounts equal to eachother and solve for the unknown value, PMT .
\[\begin{array}{l}
\ 15,000\left(1+\dfrac{.09}{12}\right)^{60}=PMT\frac{\left[\left(1+\dfrac{.09}{12}\right)^{60}-1\right]}{\dfrac{.09}{12}} \\ \ 15,000(1.5657)=PMT(75.4241) \\
\ 311.3792=PMT
\end{array} \nonumber\]
Therefore, the monthly payment needed to repay the loan is $311.38 for five years.
The formula used above (and restated here), for finding payments on an amortized loan, can appear cumbersome.
\(\begin{array}{l}
\ P\left(1+\dfrac{r}{m}\right)^{m t}=PMT\frac{\left[\left(1+\dfrac{r}{m}\right)^{m t}-1\right]}{\dfrac{r}{m}}
\end{array} \nonumber\)
If we do the necessary algebra to solve this equation for PMT , we can use the new formula to find the payments. The algebra has been omitted and the new formula is stated in the box below.
\(PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]}\)
\(P\) is the balance in the account at the beginning (the principal, or amount of the loan)
\(r\) is the annual interest rate in decimal form
\(t\) is the length of the loan, in years
\(m\) is the number of compounding periods in one year
\(PMT\) is the loan payment (monthly payment, annual payment, etc.)
*Notes:
The compounding frequency is not always explicitly given, but is determined by how often you make payments.
We will round payments on a loan
up
to the next cent.
You want to take out a $340,000 mortgage (home loan). The interest rate on the loan is 3.5%, and the loan is for 30 years. How much will your monthly payments be? How much interest will you pay over the life of the loan?
Solution
We’re looking for \(PMT\).
\(\begin{array}{ll} P = \$340,000 & \text{the starting loan amount} \\ r = 0.035 & 3.5\% \text{ annual rate} \\ t = 30 & \text{since we’re making monthly payments for 30 years} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)
\[\begin{array}{l}
\ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\
\ PMT=340000\cdot \dfrac{\left(\dfrac{0.035}{12}\right)}{\left[1-\left( 1+\dfrac{0.035}{12}\right)^{-12(30)}\right]} \\
\ PMT=1526.7519
\end{array} \nonumber\]
You will make payments of $1526.76 per month for 30 years. (Remember to round up for payments.)
You will pay a total of \(\$ 1526.76\) per month for 360 months which equals \(\$ 549,633.60\) to the loan company.
The total paid over the life of the loan is \(\$ 549,633.60 - \$ 340,000=\$ 209,633.60\).
Jack goes to a car dealer to buy a new car for $18,000 at 2% APR with a five-year loan. The dealer quotes him a monthly payment of $425. Verify that this is the correct monthly payment.
P = $18,000, r = 0.02, t = 5, m = 12
\[\begin{array}{l}
\ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\
\ PMT=18000\cdot \dfrac{\left(\dfrac{0.02}{12}\right)}{\left[1-\left( 1+\dfrac{0.02}{12}\right)^{-12(5)}\right]} \\
\ PMT=315.4996
\end{array} \nonumber\]
Jack should have a monthly payment of $315.50, not $425.
What should the total principal and interest be with the $315.50 monthly payment? \(\$ 315.50(12)(5) =\$ 18,930\)
The $315.50 payment per month has a total of $930 in interest paid over the 2-year loan period.
However, the dealer is trying to get Jack to pay $425 per month. This equates to \(\$ 425(12)(5) =\$ 25,500\) which is significantly more than the calculation above. And notice the difference in the the total interest charges; $25,500 - $18,000 = $7,500. This means that the quoted rate of 2% APR is not accurate, or the quoted price of $18,000 is not accurate, or both.
Janine bought $3,000 of new furniture on credit. Because her credit score isn’t very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month? What is the total interest charged?
- Answer
-
\(\begin{array}{ll} P = 3,000 & \text{the starting loan amount \$3,000 loan} & \\ r = 0.16 & 16\% \text{ annual rate} \\ t = 2 & \text{2 year to repay} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)
\[\begin{array}{l}
\ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\
\ PMT=3000\cdot \dfrac{\left(\dfrac{0.16}{12}\right)}{\left[1-\left( 1+\dfrac{0.16}{12}\right)^{-12(2)}\right]} \\
\ PMT=146.8893
\end{array} \nonumber\]Janine will need to make monthly payments of \(\$ 146.89\).
In total, she will pay \(\$ 3,525.36\) to the store, meaning she will pay \(\$ 525.36\) in interest over the two years.
With a fixed rate mortgage, you are guaranteed that the interest rate will not change over the life of the loan. Suppose you need $250,000 to buy a new home. The mortgage company offers you two choices: a 30-year loan with an APR of 6% or a 15-year loan with an APR of 5.5%. Compare your monthly payments and total loan cost to decide which loan you should take. Assume no difference in closing costs.
Option 1 : First calculate the monthly payment:
The monthly payment for a 30-year loan at 6% interest is $1498.88.
Now calculate the total cost of the loan over the 30 years:
The monthly payments are $1498.88 and the total cost of the loan is $539,596.80.
Option 2 : First calculate the monthly payment:
The monthly payment for a 15-year loan at 5.5% interest is $2042.71.
Now calculate the total cost of the loan over the 15 years:
The monthly payments are $2042.71 and the total cost of the loan is $367,687.80.
Therefore, the monthly payments are higher with the 15-year loan, but you spend a lot less money overall.
Present Value of an Annuity
The present valu e of an annuity is the amount of money we would need now in order to be able to make the annuity payments in the future. Often, we know how much we can afford to pay for each regular payment, so we need to find how much money we can borrow.
Jordan can afford $400 per month as a car payment. The car dealership offers an auto loan at 12% interest for 4 years. What is the present value of the car? In other words, what loan amount can Jordan afford at $400 per month?
Solution
In this example,
\(\begin{array}{ll} PMT = \$400 & \text{the monthly loan payment} \\ r = 0.12 & 12\% \text{ annual rate} \\ t = 4 & \text{since we’re making monthly payments for 4 years} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)
\[\begin{array}{l}
\ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\
\ 400=P\cdot \dfrac{\left(\dfrac{0.12}{12}\right)}{\left[1-\left( 1+\dfrac{0.12}{12}\right)^{-12(4)}\right]} \\
\ P=15189.5838
\end{array} \nonumber\]
Jordan will pay a total of $15,189.58 ($400 per month for 48 months) to the car dealership. The difference between the amount paid and the amount of the loan is the interest paid . In this case, Jordan is paying \(\$ 19,200 -\$ 15,189.58=\$ 4,010.42\) interest total.
Solve the Amortization Formula for P, Present Value
In the example above, notice we were given the monthly payment and asked to find the the present value, P . It would be helpful to solve the amortization formula for the present value first.
\(PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]}\)
---Since P is multiplied by a fraction, solve for P by multiplying both sides of the equation by the reciprocal of that fraction.
\(PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]}\cdot {\dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}}\)
---Simply the right side of the equation, then rewrite as,
\(P=PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}\)
Use this formula when you know the payment and you want to find the present value, P .
\(P=PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}\)
\(P\) is the balance in the account at the beginning (the principal, or amount of the loan)
\(r\) is the annual interest rate in decimal form
\(t\) is the length of the loan, in years
\(m\) is the number of compounding periods in one year
\(PMT\) is the loan payment (monthly payment, annual payment, etc.)
*Note: The compounding frequency is not always explicitly given, but is determined by how often you make payments.
Grace buys an iPad from a rent-to-own business on credit with payments of $30 a month for four years at 14.5% APR compounded monthly. If Grace had bought the iPad from Best Buy or Amazon it would have cost $500. What is the price that Grace paid for the iPad at the rent-to-own business? How much interest was paid over the life of the loan? What is the better option?
PMT = $30, r = 0.145, t = 4, m = 12
\[\begin{array}{l}
\ P=PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)} \\
\ P=30\cdot \dfrac{\left[1-\left(1+\dfrac{0.145}{12}\right)^{-12(4)}\right]}{\left(\dfrac{0.145}{12}\right)} \\
\ P=1087.8254
\end{array} \nonumber\]
The price Grace paid for the iPad was $1,087.83.
That’s a lot more that $500!
Also, the total amount paid over the course of the loan was . Therefore, the total amount of interest paid was $1440 - $1087.83 = $352.17.
Suppose you have won a lottery that pays $1,000 per month for the next 20 years. But, you prefer to have the entire amount now. If the interest rate is 8%, how much will you accept?
Solution
This classic present value problem needs our complete attention because the rationalization we use to solve this problem will be used again in the problems to follow.Consider, for argument purposes, that two people Mr. Cash, and Mr. Credit have won the same lottery of $1,000 per month for the next 20 years. Mr. Credit is happy with his $1,000 monthly payment, but Mr. Cash wants to have the entire amount now.
Our job is to determine how much Mr. Cash should get. We reason as follows:
If Mr. Cash accepts P dollars, then the P dollars deposited at 8% for 20 years should yield the same amount as the $1,000 monthly payments for 20 years. In other words, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like the future values to equal.
Since Mr. Cash is receiving a lump sum of \(x\) dollars, its future value is given by the lump sum formula we studied in Section 6.2, and it is
\[\mathrm{A}=\mathrm{P}(1+.08 / 12)^{240} \nonumber\]
Since Mr. Credit is receiving a sequence of payments, or an annuity, of $1,000 per month, its future value is given by the annuity formula we learned in Section 6.3. This value is
\[\mathrm{A}=\frac{\$ 1000\left[(1+.08 / 12)^{240}-1\right]}{.08 / 12} \nonumber\]
The only way Mr. Cash will agree to the amount he receives is if these two future values are equal. So we set them equal and solve for the unknown.
\[\begin{array}{l}
\mathrm{P}(1+.08 / 12)^{240}=\frac{\$ 1000\left[(1+.08 / 12)^{240}-1\right]}{.08 / 12} \\
\mathrm{P}(4.9268)=\$ 1000(589.02041) \\
\mathrm{P}(4.9268)=\$ 589020.41 \\
\mathrm{P}=\$ 119,554.36
\end{array} \nonumber\]
The present value of an ordinary annuity of $1,000 each month for 20 years at 8% is $119,554.36
The reader should also note that if Mr. Cash takes his lump sum of \(\mathrm{P}\) = $119,554.36 and invests it at 8% compounded monthly, he will have an accumulated value of \(\mathrm{A}\)=$589,020.41 in 20 years.
Which equation to use?
When presented with a finance problem (on an exam or in real life), you're usually not told what type of problem it is or which equation to use. Here are some hints on deciding which equation to use based on the wording of the problem.
The easiest types of problem to identify are loans. Loan problems almost always include words like: "loan", "amortize" (the fancy word for loans), "finance (a car)", or "mortgage" (a home loan). Look for these words. If they're there, you're probably looking at a loan problem. To make sure, see if you're given what your monthly (or annual) payment is, or if you're trying to find a monthly payment.
If the problem is not a loan, the next question you want to ask is: "Am I putting money in an account and letting it sit, or am I making regular (monthly/annually/quarterly) payments or withdrawals?" If you're letting the money sit in the account with nothing but interest changing the balance, then you're looking at a compound interest problem. The exception would be bonds and other investments where the interest is not reinvested; in those cases you’re looking at simple interest.
If you're making regular payments or withdrawals, the next questions is: "Am I putting money into the account, or am I pulling money out?" If you're putting money into the account on a regular basis (monthly/annually/quarterly) then you're looking at a basic Annuity problem. Basic annuities are when you are saving money. Usually in an annuity problem, your account starts empty, and has money in the future.
If you're pulling money out of the account on a regular basis, then you're looking at a Payout Annuity problem.
Remember, the most important part of answering any kind of question, money or otherwise, is first to correctly identify what the question is really asking, and to determine what approach will best allow you to solve the problem.
For each of the following scenarios, determine if it is a compound interest problem, a savings annuity problem, a payout annuity problem, or a loans problem. Then solve each problem.
- Marcy received an inheritance of $20,000, and invested it at 6% interest. She is going to use it for college, withdrawing money for tuition and expenses each quarter. How much can she take out each quarter if she has 3 years of school left?
- Paul wants to buy a new car. Rather than take out a loan, he decides to save $200 a month in an account earning 3% interest compounded monthly. How much will he have saved up after 3 years?
- Keisha is managing investments for a non-profit company. They want to invest some money in an account earning 5% interest compounded annually with the goal to have $30,000 in the account in 6 years. How much should Keisha deposit into the account?
- Miao is going to finance new office equipment at a 2% rate over a 4 year term. If she can afford monthly payments of $100, how much new equipment can she buy?
- How much would you need to save every month in an account earning 4% interest to have $5,000 saved up in two years?
- Answer
-
- This is a payout annuity problem. She can pull out $1833.60 a quarter.
- This is a savings annuity problem. He will have saved up $7,524.11.
- This is compound interest problem. She would need to deposit $22,386.46.
- This is a loans problem. She can buy $4,609.33 of new equipment.
- This is a savings annuity problem. You would need to save $200.46 each month


2.1.7: Free Variables and Sentences
The free occurrences of a variable in a formula are defined inductively as follows:
- \(\indcaseA{A}{$A$ is atomic}\) all variable occurrences in \(\indfrm\) are free.
- \(\indcase{A}{\lnot B}\) the free variable occurrences of \(\indfrm\) are exactly those of \(B\) .
- \(\indcase{A}{(B \ast C)}\) the free variable occurrences of \(\indfrm\) are those in \(B\) together with those in \(C\) .
- \(\indcase{A}{\lforall{x}{B}}\) the free variable occurrences in \(\indfrm\) are all of those in \(B\) except for occurrences of \(x\) .
- \(\indcase{A}{\lexists{x}{B}}\) the free variable occurrences in \(\indfrm\) are all of those in \(B\) except for occurrences of \(x\) .
An occurrence of a variable in a formula \(A\) is bound if it is not free.
Give an inductive definition of the bound variable occurrences along the lines of Definition \(\PageIndex{1}\) .
If \(\lforall{x}{B}\) is an occurrence of a subformula in a formula \(A\) , then the corresponding occurrence of \(B\) in \(A\) is called the scope of the corresponding occurrence of \(\lforall{x}{}\) . Similarly for \(\lexists{x}{}\) .
If \(B\) is the scope of a quantifier occurrence \(\lforall{x}{}\) or \(\lexists{x}{}\) in \(A\) , then the free occurrences of \(x\) in \(B\) are bound in \(\lforall{x}{B}\) and \(\lexists{x}{B}\) . We say that these occurrences are bound by the mentioned quantifier occurrence.
Consider the following formula: \[\lexists{\Obj v_0}{\underbrace{\Atom{\Obj A^2_0}{\Obj v_0,\Obj v_1}}_{B}}\nonumber\] \(B\) represents the scope of \(\lexists{\Obj v_0}{}\) . The quantifier binds the occurence of \(\Obj v_0\) in \(B\) , but does not bind the occurence of \(\Obj v_1\) . So \(\Obj v_1\) is a free variable in this case.
We can now see how this might work in a more complicated formula \(A\) : \[\lforall{\Obj v_0}{\underbrace{(\Atom{\Obj A^1_0}{\Obj v_0} \lif \Atom{\Obj A^2_0}{\Obj v_0, \Obj v_1})}_{B}} \lif \lexists{\Obj v_1}{\underbrace{(\Atom{\Obj A^2_1}{\Obj v_0, \Obj v_1} \lor \lforall{\Obj v_0}{\overbrace{\lnot \Atom{\Obj A^1_1}{\Obj v_0}}^{D}})}_{C}}\nonumber\] \(B\) is the scope of the first \(\lforall{\Obj v_0}{}\) , \(C\) is the scope of \(\lexists{\Obj v_1}{}\) , and \(D\) is the scope of the second \(\lforall{\Obj v_0}{}\) . The first \(\lforall{\Obj v_0}{}\) binds the occurrences of \(\Obj v_0\) in \(B\) , \(\lexists{\Obj v_1}{}\) the occurrence of \(\Obj v_1\) in \(C\) , and the second \(\lforall{\Obj v_0}{}\) binds the occurrence of \(\Obj v_0\) in \(D\) . The first occurrence of \(\Obj v_1\) and the fourth occurrence of \(\Obj v_0\) are free in \(A\) . The last occurrence of \(\Obj v_0\) is free in \(D\) , but bound in \(C\) and \(A\) .
A formula \(A\) is a sentence iff it contains no free occurrences of variables.


4.2: Definition of Exponential Function, Exponential Growth and Decay Models
In this section, you will learn to
- Define an exponential function
- Recognize and model exponential growth and decay
- Compare linear and exponential growth
- Distinguish between exponential and power functions
A line is of the form y = ax + b Doman: (-\(\infty, \infty\)) Range : (-\(\infty, \infty\))
A quadratic is of the form y = ax 2 + bx + c Domain: (-\(\infty, \infty\)) Range: depends on the values of a, b, and c. If a > 0, the Range is [-b/2a, \(\infty\)). If a < 0, the Range is (-\(\infty\), -b/2a]
A cubic is of the form y = ax 3 + bx 2 + cx + d Domain: (-\(\infty, \infty\)) Range: (-\(\infty, \infty\))
An exponential function is different from the function above because the x variable is in the exponent. The exponential function takes the following form.
y = ab x . where b > 0. Domain: (-\(\infty, \infty\)) Range: If a > 0, y > 0 or (0, \(\infty\)).
If b > 0, the exponential function increases as x increases. For example y = 2 x . The leading coefficient a = 1.
If 0 < b < 1, the exponential function decreases as x increases. For example y = (1/2) x . The leading coefficient a = 1.
Comparing Exponential and Linear Growth
Consider two social media sites which are expanding the number of users they have:
- Site A has 10,000 users, and expands by adding 1,500 new users each month
- Site B has 10,000 users, and expands by increasing the number of users by 10% each month.
The number of users for Site A can be modeled as linear growt h. The number of users increases by a constant number, 1500, each month. If \(x\) = the number of months that have passed and \(y\) is the number of users, the number of users after \(x\) months is \(y = 10000+1500x\). For site B, the user base expands by a constant percent each month , rather than by a constant number. Growth that occurs at a constant percent each unit of time is called exponential growth.
We can look at growth for each site to understand the difference. The table compares the number of users for each site for 12 months. The table shows the calculations for the first 4 months only, but uses the same calculation process to complete the rest of the 12 months.
| Month | Users at Site A | Users at Site B |
|---|---|---|
| 0 | \(10000\) | \(10000\) |
| 1 | \(10000 + 1500 = 11500\) |
\(\begin{aligned}
10000+&10 \% \text { of } 10000 \\ =& 10000+0.10(10000) \\ =& 10000(1.10)=11000 \end{aligned}\) |
| 2 | \(11500 + 1500 = 13000\) |
\(\begin{aligned}
|
| 3 | \(13000 + 1500 = 14500\) |
\(\begin{aligned}
|
| 4 | \(14500 + 1500 = 16000\) |
\(\begin{aligned}
|
| 5 | \(17500\) | \(16105\) |
| 6 | \(19000\) | \(17716\) |
| 7 | \(20500\) | \(19487\) |
| 8 | \(22000\) | \(21436\) |
| 9 | \(23500\) | \(23579\) |
| 10 | \(25000\) | \(25937\) |
| 11 | \(26500\) | \(28531\) |
| 12 | \(28000\) | \(31384\) |
For Site B, we can re-express the calculations to help us observe the patterns and develop a formula for the number of users after x months.
- Month 1: \(y = 10000(1.1) = 11000\)
- Month 2: \(y = 11000(1.1) = 10000(1.1)(1.1)=\mathbf{10000(1.1)^2}= 12100\)
- Month 3: \(y = 12100(1.1) = 10000(1.1)^2 (1.1)=\mathbf{10000(1.1)^3}= 13310\)
- Month 4: \(y = 13310(1.1) = 10000(1.1)^3 (1.1)=\mathbf{10000(1.1)^4}= 14641\)
By looking at the patterns in the calculations for months 2, 3, and 4, we can generalize the formula. After \(x\) months, the number of users \(y\) is given by the function \(\mathbf{y = 10000(1.1)^x}\)
Using Exponential Functions to Model Growth and Decay
In exponential growth , the value of the dependent variable \(y\) increases at a constant percentage rate as the value of the independent variable (\(x\) or \(t\)) increases. Examples of exponential growth functions include:
- the number of residents of a city or nation that grows at a constant percent rate.
- the amount of money in a bank account that earns interest if money is deposited at a single point in time and left in the bank to compound without any withdrawals.
In exponential decay, the value of the dependent variable y decreases at a constant percentage rate as the value of the independent variable (\(x\) or \(t\)) increases. Examples of exponential decay functions include:
- value of a car or equipment that depreciates at a constant percent rate over time
- the amount a drug that still remains in the body as time passes after it is ingested
- the amount of radioactive material remaining over time as a radioactive substance decays.
Exponential functions often model quantities as a function of time; thus we often use the letter \(t\) as the independent variable instead of \(x\).
The table compares exponential growth and exponential decay functions:
| Exponential Growth | Exponential Decay |
|---|---|
|
Quantity grows by a constant percent
|
Quantity decreases by a constant percent per unit of time |
|
\(\mathbf{y=ab^x}\)
|
\(\mathbf{y=ab^x}\)
|
In general, the domain of exponential functions is the set of all real numbers. The range of an exponential growth or decay function is the set of all positive real numbers.
In most applications, the independent variable, \(x\) or \(t\), represents time. When the independent variable represents time, we may choose to restrict the domain so that independent variable can have only non-negative values in order for the application to make sense. If we restrict the domain, then the range is also restricted as well.
- For an exponential growth function \(y=ab^x\) with \(b>1\) and \(a > 0\), if we restrict the domain so that \(x ≥ 0\), then the range is \(y ≥ a\).
- For an exponential decay function \(y=ab^x\) with \(0<b<1\) and \(a > 0\), if we restrict the domain so that \(x ≥ 0\), then the range is \(0 < y ≤ a\).
Consider the growth models for social media sites A and B, where \(x\) = number of months since the site was started and \(y\) = number of users. The number of users for Site A follows the linear growth model:
\[y = 10000+1500x \nonumber. \nonumber \]
The number of users for Site B follows the exponential growth model:
\[y=10000(1.1^x) \nonumber \]
For each site, use the function to calculate the number of users at the end of the first year, to verify the values in the table. Then use the functions to predict the number of users after 30 months.
Solution
Since \(x\) is measured in months, then \(x = 12\) at the end of one year.
Linear Growth Model:
When \(x = 12\) months, then \(y = 10000 + 1500(12) = 28,000\) users
When \(x = 30\) months, then \(y = 10000 + 1500(30) = 55,000\) users
Exponential Growth Model:
When \(x = 12\) months, then \(y = 10000(1.1^{12}) = 31,384\) users
When \(x = 30\) months, then \(y = 10000(1.1^{30}) =174,494\) users
We see that as \(x\), the number of months, gets larger, the exponential growth function grows large faster than the linear function (even though in Example \(\PageIndex{1}\) the linear function initially grew faster). This is an important characteristic of exponential growth: exponential growth functions always grow faster and larger in the long run than linear growth functions.
It is helpful to use function notation, writing \(y = f(t) = ab^t\), to specify the value of \(t\) at which the function is evaluated.
A forest has a population of 2000 squirrels that is increasing at the rate of 3% per year. Let \(t\) = number of years and \(y = f(t) =\) number of squirrels at time \(t\).
- Find the exponential growth function that models the number of squirrels in the forest at the end of \(t\) years.
- Use the function to find the number of squirrels after 5 years and after 10 years
Solution
a. The exponential growth function is \(y = f(t) = ab^t\), where \(a = 2000\) because the initial population is 2000 squirrels
The annual growth rate is 3% per year, stated in the problem. We will express this in decimal form as \(r = 0.03\)
Then \(b = 1+r = 1+0.03 = 1.03\)
Answer: The exponential growth function is \(y = f(t) = 2000(1.03^t)\)
b. After 5 years, the squirrel population is \(y = f(5) = 2000(1.03^5) \approx 2319\) squirrels
After 10 years, the squirrel population is \(y = f(10) = 2000(1.03^{10}) \approx 2688\) squirrels
A large lake has a population of 1000 frogs. Unfortunately the frog population is decreasing at the rate of 5% per year. Let \(t\) = number of years and \(y\) = \(g(t)\) = the number of frogs in the lake at time \(t\).
- Find the exponential decay function that models the population of frogs.
- Calculate the size of the frog population after 10 years.
Solution
a. The exponential decay function is \(y = g(t) = ab^t\), where \(a = 1000\) because the initial population is 1000 frogs
The annual decay rate is 5% per year, stated in the problem. The words decrease and decay indicated that \(r\) is negative. We express this as \(r = -0.05\) in decimal form.
Then, \(b = 1+ r = 1+ (-0.05) = 0.95\)
Answer: The exponential decay function is: \(y = g(t) = 1000(0.95^t)\)
b. After 10 years, the frog population is \(y = g(10) = 1000(0.95^{10}) \approx 599\) frogs
A population of bacteria is given by the function \(y = f(t) = 100(2^t)\), where \(t\) is time measured in hours and \(y\) is the number of bacteria in the population.
- What is the initial population?
- What happens to the population in the first hour?
- How long does it take for the population to reach 800 bacteria?
Solution
a. The initial population is 100 bacteria. We know this because \(a = 100\) and because at time \(t = 0\), then \(f(0) = 100(2^0) = 100(1)=100\)
b. At the end of 1 hour, the population is \(y = f(1) = 100(2^1) = 100(2)=200\) bacteria.
The population has doubled during the first hour.
c. We need to find the time \(t\) at which \(f(t) = 800\). Substitute 800 as the value of \(y\):
\[\begin{array}{l}
y=f(t)=100\left(2^{t}\right) \\
800=100\left(2^{t}\right)
\end{array}\nonumber \]
Divide both sides by 100 to isolate the exponential expression on the one side
\[8=1\left(2^{\mathrm{t}}\right) \nonumber \]
8 = 2 3 , so it takes \(t = 3\) hours for the population to reach 800 bacteria.
Two important notes about Example \(\PageIndex{4}\):
- In solving \(8 = 2^t\), we “knew” that \(t\) is 3. But we usually can not know the value of the variable just by looking at the equation. Later we will use logarithms to solve equations that have the variable in the exponent.
- To solve \(800 = 100(2^t)\), we divided both sides by 100 to isolate the exponential expression \(2^t\). We can not multiply 100 by 2. Even if we write it as \(800 =100(2)^t\), which is equivalent, we still can not multiply 100 by 2. The exponent applies only to the quantity immediately before it, so the exponent t applies only to the base of 2.
Comparing Linear, Exponential and Power Functions
To identify the type of function from its formula, we need to carefully note the position that the variable occupies in the formula.
A linear function can be written in the form \(\mathbf{y=a x+b}\)
As we studied in chapter 1, there are other forms in which linear equations can be written, but linear functions can all be rearranged to have form \(y = mx + b\).
An exponential function has form \(\mathbf{y=ab^x}\)
The variable \(\mathbf{x}\) is in the exponent. The base \(b\) is a positive number.
- If \(b>1\), the function represents exponential growth.
- If \(0 < b < 1\), the function represents exponential decay
A power function has form \(\mathbf{y=cx^P}\)
The variable \(\mathbf{x}\) is in the base. The exponent \(p\) is a non-zero number.
We compare three functions:
- linear function \(y = f(x) = 2x\)
- exponential function \(y = g(x) = 2^x\)
- power function \(y = h(x) = x^2\)
| \(\mathbf{x}\) | \(\mathbf{y = f(x) =2x}\) | \(\mathbf{y = g(x) =2^x}\) | \(\mathbf{y= h(x) =x^2}\) |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 1 | 2 | 2 | 1 |
| 2 | 4 | 4 | 4 |
| 3 | 6 | 8 | 9 |
| 4 | 8 | 16 | 16 |
| 5 | 10 | 32 | 25 |
| 6 | 12 | 64 | 36 |
| 10 | 20 | 1024 | 100 |
| Type of function | Linear \(y = mx+b\) | Exponential \(y = ab^x\) | Power \(y = cx^P\) |
|
How to recognize this type of
function. |
all terms are first degree; \(m\) is slope;
|
base is a number \(b>0\); the variable
|
variable is in the base; exponent is a number \(\mathrm{p} \neq 0\) |
|
For equal intervals of change in
\(x\), \(y\) increases by a constant amount |
For equal intervals of change in
\(x\), \(y\) increases by a constant ratio |
For the functions in the previous table: linear function \(y = f(x) = 2x\), exponential function \(y = g(x) = 2^x\), and power function \(y = h(x) = x^2\), if we restrict the domain to \(x ≥ 0\) only, then all these functions are growth functions. When \(x ≥ 0\), the value of \(y\) increases as the value of \(x\) increases.
The exponential growth function grows large faster than the linear and power functions, as \(x\) gets large. This is always true of exponential growth functions, as \(x\) gets large enough.
Classify the functions below as exponential, linear, or power functions.
- \(y=10x^3\)
- \(y=1000-30x\)
- \(y=1000\left(1.05^x\right)\)
- \(y=500(0.75^x)\)
- \(y=10\sqrt[3]{x}=x^{1/3}\)
- \(y=5x-1\)
- \(y=6/x^2=6x^{-2}\)
Solution:
The exponential functions are
c. \(y=1000\left(1.05^x\right)\) The variable is in the exponent; the base is the number \(b = 1.05\)
d. \(y=500(0.75^x)\)\) The variable is in the exponent; the base is the number \(b = 0.75\)
The linear functions are
b. \(y=1000-30x\)
f. \(y=5x-1\)
The power functions are
a. \(y=10x^3\) The variable is the base; the exponent is a fixed number, \(p=3\).
e. \(y=10\sqrt[3]{x}=x^{1/3}\) The variable is the base; the exponent is a number, \(p=1/3\).
g. \(y=6/x^2=6x^{-2}\) The variable is the base; the exponent is a number, \(p = -2\).
NATURAL BASE: e
The number e is often used as the base of an exponential function. e is called the natural base.
e is approximately 2.71828
e is an irrational number with an infinite number of decimals; the decimal pattern never repeats.
Section 6.2 includes an example that shows how the value of e is developed and why this number is mathematically important. Students studying Finite Math should already be familiar with the number e from their prerequisite algebra classes.
When e is the base in an exponential growth or decay function, it is referred to as continuous growth or continuous decay . We will use e in chapter 6 in financial calculations when we examine interest that compounds continuously.
Any exponential function can be written in the form \(\mathbf{y = ae^{kx}}\)
\(\mathbf{k}\) is called the continuous growth or decay rate.
- If \(k > 0\), the function represents exponential growth
- If \(k< 0\), the function represents exponential decay
\(\mathbf{a}\) is the initial value
We can rewrite the function in the form \(\mathbf{y = ab^x}\) , where \(\mathbf{b=e^k}\)
In general, if we know one form of the equation, we can find the other forms. For now, we have not yet covered the skills to find \(k\) when we know \(b\). After we learn about logarithms later in this chapter, we will find \(k\) using natural log: \(k = \ln b\).
The table below summarizes the forms of exponential growth and decay functions.
| y = ab x | y = a(1+r) x | y = a e kx , k ≠ 0 | |
|---|---|---|---|
| Initial value | a>0 | a>0 | a>0 |
| Relationship between b, r, k | b > 0 | b=1+ r | b = e k and k = ln b |
| Growth | b > 1 | r > 0 | k > 0 |
| Decay | 0 < b < 1 | r < 0 | k < 0 |
The value of houses in a city are increasing at a continuous growth rate of 6% per year. For a house that currently costs $400,000:
- Write the exponential growth function in the form \(y=ae^{kx}\).
- What would be the value of this house 4 years from now?
- Rewrite the exponential growth function in the form \(y=ab^x\).
- Find and interpret \(r\).
Solution
a. The initial value of the house is \(a\) = $400000
The problem states that the continuous growth rate is 6% per year, so \(k\) = 0.06
The growth function is : \(y=400000e^{0.06x}\)
b. After 4 years, the value of the house is \(y=400000e^{0.06 (4)}\) = $508,500.
c. To rewrite \(y=400000e^{0.06x}\) in the form \(y = ab^x\), we use the fact that \(b=e^k\).
\[\begin{array}{l}
\mathrm{b}=e^{0.06} \\
\mathrm{b}=1.06183657 \approx 1.0618 \\
\mathrm{y}=400000(1.0618)^{\mathrm{x}}
\end{array}\nonumber \]
d. To find \(r\), we use the fact that \(b=1+ r\)
\[\begin{array}{l}
\mathrm{b}=1.0618 \\
1+\mathrm{r}=1.0618 \\
\mathrm{r}=0.0618
\end{array} \nonumber \]
The value of the house is increasing at an annual rate of 6.18%.
Suppose that the value of a certain model of new car decreases at a continuous decay rate of 8% per year. For a car that costs $20,000 when new:
- Write the exponential decay function in the form \(y=ae^{kx}\).
- What would be the value of this car 5 years from now?
- Rewrite the exponential decay function in the form \(y=ab^x\).
- Find and interpret \(r\).
Solution
a. The initial value of the car is \(a\) = $20000
The problem states that the continuous decay rate is 8% per year, so \(k\) = -0.08
The growth function is : \(y=20000e^{-0.08x}\)
b. After 5 years, the value of the car is \(y=20000 e^{-0.08 (5)}\) = $13,406.40.
c. To rewrite \(y=20000e^{-0.08x}\) in the form \(y=ab^x\), we use the fact that \(b=e^k\).
\[\begin{array}{l}
\mathrm{b}=e^{-0.08} \\
\mathrm{b}=0.9231163464 \approx 0.9231 \\
\mathrm{y}=20000(0.9231)^{\mathrm{x}}
\end{array} \nonumber \]
d. To find \(r\), we use the fact that \(b=1+ r\)
\[\begin{array}{l}
\mathrm{b}=0.9231 \\
\mathrm{l}+\mathrm{r}=0.9231 \\
\mathrm{r}=0.9231-1=-0.0769
\end{array} \nonumber \]
The value of the car is decreasing at an annual rate of 7.69%.