2.4: Newton's Second Law of Motion- Force and Acceleration
Learning Objectives
- Define net force, external force, and system.
- Understand Newton’s second law of motion.
- Apply Newton’s second law to determine the weight of an object.
Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned.
First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is an acceleration . Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration .
Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct—an external force acts from outside the system of interest. For example, in Figure \(\PageIndex{1}\)(a) the system of interest is the wagon plus the child in it. The two forces exerted by the other children are external forces. An internal force acts between elements of the system. Again looking at Figure \(\PageIndex{1}\)(a), the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. (The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of the system before you can determine which forces are external . Sometimes the system is obvious, whereas other times identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept will be revisited many times on our journey through physics.
Figure \(\PageIndex{1}\) is our first example of a free-body diagram , which is a technique used to illustrate all the external forces acting on a body. The body is represented by a single isolated point (or free body), and only those forces acting on the body from the outside (external forces) are shown. Free-body diagrams are very useful in analyzing forces acting on a system and are employed extensively in the study and application of Newton’s laws of motion.
Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure \(\PageIndex{1}\). In part (a), a smaller force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight \(w\) and the support of the ground \(N\), and the horizontal force \(f\) represents the force of friction. These will be discussed in more detail in later sections. For now, we will say that friction is a force that opposes the motion past each other of objects that are touching. Figure \(\PageIndex{1}\)(b) shows how vectors representing the external forces add together to produce a net force, \(\boldsymbol{F}_{\text {net }}\).
To obtain an equation for Newton’s second law, we first write the relationship of acceleration and net external force as the proportionality
\[\boldsymbol{a} \propto \boldsymbol{F}_{\text {net }} \nonumber \]
where the symbol ∝ means “proportional to,” and \(\boldsymbol{F}_{\text {net }}\) is the net external force , the vector sum of all external forces. This proportionality states what we have said in words— acceleration is directly proportional to the net external force . Once the system of interest is chosen, it is important to identify the external forces, so that the internal forces can be ignored (again, as we will see later, they exactly cancel out, allowing for a tremendous simplification).
Now, hopefully it also seems reasonable that acceleration should be inversely proportional to the mass of the system. That is, if the system has more “stuff”, then for the same net external force applied, the acceleration is less. And indeed, as illustrated in Figure \(\PageIndex{2}\), the same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The proportionality is written as
\[a \propto \frac{1}{m} \nonumber \]
where \(m\) is the mass of the system. In fact, this is how we will define mass. That is, for an object B to have double the mass of an object A, the same net external force applied on object B produces half the acceleration it does on object A.
As it turns out, the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton's second law of motion.
NEWTON’S SECOND LAW OF MOTION
The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.
In equation form, Newton’s second law of motion is
\[\boldsymbol{a}=\frac{\boldsymbol{F}_{\text {net }}}{m} \nonumber \]
This is often written in the more familiar form
\[\boldsymbol{F}_{\text {net }}=m \boldsymbol{a}, \nonumber \]
with the vector notation indicating that the net external force is in the same direction as acceleration. When only the magnitude of force and acceleration are considered, this equation is simply (note the lack of vector notations)
\[F_{\text {net }}=m a . \nonumber \]
Although these equations are really the same, the first gives more insight into what Newton’s second law means. The law is a cause and effect relationship among three quantities. That is, acceleration is caused by a net external force, not vice versa, as the second equation might mistakenly imply.
Units of Force
\(F_{\text {net }}=m a\) is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is called the newton (abbreviated N) and 1 N is the force needed to accelerate a system of mass 1 kg at the rate of \(1 \mathrm{~m} / \mathrm{s}^{2}\). Putting these together,
\[1 \mathrm{~N}=1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}. \nonumber \]
While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb.
Weight and the Gravitational Force
When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight , \(w\). Weight can be denoted as a vector \(w\) because it has a direction; down is, by definition, the direction of gravity, so \(w\) is directed downward. The magnitude of weight is denoted as \(w\). Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration \(g\). Using Galileo’s result and Newton’s second law, we can derive an equation for the magnitude of weight.
Consider an object with mass \(m\) falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude \(w\). Newton’s second law states that the magnitude of the net external force on an object is \(F_{\text {net }}=m a \).
Since the object experiences only the downward force of gravity, \(F_{\text {net }}=w\). We know that the acceleration of an object due to gravity is \(g\), or \(a=g\). Substituting these into Newton’s second law gives
Definition: WEIGHT
This is the equation for weight —the gravitational force on a mass \(m\):
\[w=m g. \nonumber \]
Since \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\) on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see:
\[w=m g=(1.0 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)=9.8 \mathrm{~N}. \nonumber \]
When the net external force on an object is its weight, we say that it is in free-fall . That is, the only force acting on the object is the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air acting on the object.
The acceleration due to gravity \(g\) varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth. On the Moon, for example, the acceleration due to gravity is only \(1.67 \mathrm{~m} / \mathrm{s}^{2}\). A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon.
The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body , such as Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of “weightlessness” and “microgravity,” they are really referring to the phenomenon we call “free-fall” in physics. We shall use the above definition of weight, and we will make careful distinctions between free-fall and actual weightlessness.
It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter (how much “stuff”) and does not vary, whereas weight is the gravitational force and does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our “weight” in kilograms, but never in the correct units of newtons.
COMMON MISCONCEPTIONS: MASS VS. WEIGHT
Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly different from one another. Mass is a measure of how much matter is in an object. The typical measure of mass is the kilogram (or the “slug” in English units). Weight, on the other hand, is a measure of the force of gravity acting on an object. Weight is equal to the mass of an object (\(m\)) multiplied by the acceleration due to gravity (\(g\)). Like any other force, weight is measured in terms of newtons (or pounds in English units). This interchangeable use is why you might have heard an expression such as, “1 kilogram is 2.2 pounds.” The correct statement is “1 kilogram mass weighs 2.2 pounds on Earth.”
Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight depends on the acceleration due to gravity, the weight of an object can change when the object enters into a region with stronger or weaker gravity. For example, the acceleration due to gravity on the Moon is \(1.67 \mathrm{~m} / \mathrm{s}^{2}\) (which is much less than the acceleration due to gravity on Earth, \(9.80 \mathrm{~m} / \mathrm{s}^{2}\)). If you measured your weight on Earth and then measured your weight on the Moon, you would find that you “weigh” much less, even though you do not look any skinnier. This is because the force of gravity is weaker on the Moon. In fact, when people say that they are “losing weight,” they really mean that they are losing “mass” (which in turn causes them to weigh less).
TAKE-HOME EXPERIMENT: MASS AND WEIGHT
What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. The springs provide a measure of your weight (for an object which is not accelerating). This is a force in newtons (or pounds). In most countries, the measurement in newtons is divided by 9.80 to give a reading in mass units of kilograms. The scale measures weight but is calibrated to provide information about mass. While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? Would your scale measure the same “mass” on Earth as on the Moon?
Example \(\PageIndex{1}\): What Acceleration Can a Person Produce when Pushing a Lawn Mower?
Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the mower is 24 kg. What is its acceleration?
Strategy
Since \(F_{\text {net }}\) and \(m\) are given, the acceleration can be calculated directly from Newton’s second law as stated in \(F_{\text {net }}=ma\).
Solution
The magnitude of the acceleration \(a\) is \(a=\frac{F_{\text {net }}}{m}\). Entering known values gives
\[a=\frac{51 \mathrm{~N}}{24 \mathrm{~kg}} \nonumber\]
Substituting the units \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{2}\) for N yields
\[a=\frac{51 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}}{24 \mathrm{~kg}}=2.1 \mathrm{~m} / \mathrm{s}^{2}. \nonumber\]
Discussion
The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower forward must be greater than the friction opposing the motion (directed backward) to result in a forward net force, and the vertical forces must cancel if there is to be no acceleration in the vertical direction. The acceleration found is small enough to be reasonable for a person pushing a mower. When the person reaches the top speed, the acceleration will be zero (no change of velocity) and the force exerted by the person pushing the mower will be equal to the friction opposing the motion.
Example \(\PageIndex{2}\): What Rocket Thrust Accelerates This Sled?
Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust \(T\), for the four-rocket propulsion system shown in Figure \(\PageIndex{4}\). The sled’s initial acceleration is \(49 \mathrm{~m} / \mathrm{s}^{2}\) the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.
Strategy
Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.
Solution
Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with
\[F_{\text {net }}=m a, \nonumber\]
where \(F_{\text {net }}\) is the net force along the horizontal direction. We can see from Figure \(\PageIndex{4}\) that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is
\[F_{\text {net }}=4 T-f. \nonumber\]
Substituting this into Newton’s second law gives
\[F_{\text {net }}=m a=4 T-f. \nonumber\]
Using a little algebra, we solve for the total thrust 4 T :
\[4 T=m a+f . \nonumber\]
Substituting known values yields
\[4 T=m a+f=(2100 \mathrm{~kg})\left(49 \mathrm{~m} / \mathrm{s}^{2}\right)+650 \mathrm{~N} . \nonumber\]
So the total thrust is
\[4 T=1.04 \times 10^{5} \mathrm{~N}, \nonumber\]
and the individual thrusts are
\[T=\frac{1.04 \times 10^{5} \mathrm{~N}}{4}=2.6 \times 10^{4} \mathrm{~N}. \nonumber\]
Discussion
The numbers are quite large, so to put the result in perspective, note that acceleration of \(49 \mathrm{~m} / \mathrm{s}^{2}\) is approximately 5 times the gravitational acceleration on Earth. It's a large acceleration, which requires a large net external force to produce.
Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 \(g^{\prime}\)s. (Recall that \(g\), the acceleration due to gravity, is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\). When we say that an acceleration is \(g^{\prime}\)s, it is \(45 \times 9.80 \mathrm{~m} / \mathrm{s}^{2}\), which is approximately \(440 \mathrm{~m} / \mathrm{s}^{2}\). While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.
Newton’s second law of motion gives a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. The next section introduces the third and final law of motion.
Section Summary
- Acceleration, \(a\), is defined as a rate of change in velocity, resulting from a change in the magnitude and/or the direction of the velocity.
- An external force is one acting on a system from outside the system, as opposed to internal forces, which act between components within the system.
- Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.
- In equation form, Newton’s second law of motion is \(\boldsymbol{a}=\frac{\boldsymbol{F}_{\text {net }}}{m}\), often written in the more familiar form: \(\boldsymbol{F}_{\text {net }}=m \boldsymbol{a} \).
-
The weight \(w\) of an object is defined as the force of gravity acting on an object of mass mm. Given acceleration due to gravity \(g\), the magnitude of weight is:
\[w=m g \nonumber\]
Glossary
- acceleration
- the rate at which an object’s velocity changes over a period of time
- free-fall
- a situation in which the only force acting on an object is the force due to gravity
- external force
- a force acting on an object or system that originates outside of the object or system
- net external force
- the vector sum of all external forces acting on an object or system; causes a mass to accelerate
- free-body diagram
- a sketch showing all of the external forces acting on an object or system; the system is represented by a dot, and the forces are represented by vectors extending outward from the dot
- Newton’s second law of motion
- The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.
- system
- the object or the group of objects under consideration
- weight
- the force due to gravity; \(w=mg\) for objects on Earth


4.6: Planning
Planning is the fourth step of the nursing process (and the fourth Standard of Practice set by the American Nurses Association). This standard is defined as, “The registered nurse develops a collaborative plan encompassing strategies to achieve expected outcomes.” The RN develops an individualized, holistic, evidence-based plan in partnership with the health care consumer, family, significant others, and interprofessional team. Elements of the plan are prioritized. The plan is modified according to the ongoing assessment of the health care consumer’s response and other indicators. The plan is documented using standardized language or terminology. [1]
After expected outcomes are identified, the nurse begins planning nursing interventions to implement. Nursing interventions are evidence-based actions that the nurse performs to achieve patient outcomes. Just as a provider makes medical diagnoses and writes prescriptions to improve the patient’s medical condition, a nurse formulates nursing diagnoses and plans nursing interventions to resolve patient problems. Nursing interventions should focus on eliminating or reducing the related factors (etiology) of the nursing diagnoses when possible. [2] Nursing interventions, goals, and expected outcomes are written in the nursing care plan for continuity of care across shifts, nurses, and health professionals.
Planning Nursing Interventions
You might be asking yourself, “How do I know what evidence-based nursing interventions to include in the nursing care plan?” There are several sources that nurses and nursing students can use to select nursing interventions. Many agencies have care planning tools and references included in the electronic health record that are easily documented in the patient chart. Nurses can also refer to other care planning books our sources such as the Nursing Interventions Classification (NIC) system. Based on research and input from the nursing profession, NIC categorizes and describes nursing interventions that are constantly evaluated and updated. Interventions included in NIC are considered evidence-based nursing practices. The nurse is responsible for using clinical judgment to make decisions about which interventions are best suited to meet an individualized patient’s needs. [3]
Direct and Indirect Care
Nursing interventions are considered direct care or indirect care. Direct care refers to interventions that are carried out by having personal contact with patients. Examples of direct care interventions are wound care, repositioning, and ambulation. Indirect care interventions are performed when the nurse provides assistance in a setting other than with the patient. Examples of indirect care interventions are attending care conferences, documenting, and communicating about patient care with other providers.
Classification of Nursing Interventions
There are three types of nursing interventions: independent, dependent, and collaborative. (See Figure 4.12 [4] for an image of a nurse collaborating with the health care team when planning interventions.)
Independent Nursing Interventions
Any intervention that the nurse can independently provide without obtaining a prescription is considered an independent nursing intervention . An example of an independent nursing intervention is when the nurses monitor the patient’s 24-hour intake/output record for trends because of a risk for imbalanced fluid volume. Another example of independent nursing interventions is the therapeutic communication that a nurse uses to assist patients to cope with a new medical diagnosis.
Example. Refer to Scenario C in the “Assessment” section of this chapter. Ms. J. was diagnosed with Fluid Volume Excess . An example of an evidence-based independent nursing intervention is, “The nurse will reposition the patient with dependent edema frequently, as appropriate.” [5] The nurse would individualize this evidence-based intervention to the patient and agency policy by stating, “The nurse will reposition the patient every 2 hours.”
Dependent Nursing Interventions
Dependent nursing interventions require a prescription before they can be performed. Prescriptions are orders, interventions, remedies, or treatments ordered or directed by an authorized primary health care provider. [6] A primary health care provider is a member of the health care team (usually a physician, advanced practice nurse, or physician’s assistant) who is licensed and authorized to formulate prescriptions on behalf of the client. For example, administering medication is a dependent nursing intervention. The nurse incorporates dependent interventions into the patient’s overall care plan by associating each intervention with the appropriate nursing diagnosis.
Example. Refer to Scenario C in the “Assessment” section of this chapter. Ms. J. was diagnosed with Fluid Volume Excess . An example of a dependent nursing intervention is, “The nurse will administer scheduled diuretics as prescribed.”
Collaborative Nursing Interventions
Collaborative nursing interventions are actions that the nurse carries out in collaboration with other health team members, such as physicians, social workers, respiratory therapists, physical therapists, and occupational therapists. These actions are developed in consultation with other health care professionals and incorporate their professional viewpoint. [7]
Example. Refer to Scenario C in the “Assessment” section of this chapter. Ms. J. was diagnosed with Fluid Volume Excess . An example of a collaborative nursing intervention is consulting with a respiratory therapist when the patient has deteriorating oxygen saturation levels. The respiratory therapist plans oxygen therapy and obtains a prescription from the provider. The nurse would document “ The nurse will manage oxygen therapy in collaboration with the respiratory therapist ” in the care plan.
Individualization of Interventions
It is vital for the planned interventions to be individualized to the patient to be successful. For example, adding prune juice to the breakfast meal of a patient with constipation will only work if the patient likes to drink the prune juice. If the patient does not like prune juice, then this intervention should not be included in the care plan. Collaboration with the patient, family members, significant others, and the interprofessional team is essential for selecting effective interventions. The number of interventions included in a nursing care plan is not a hard and fast rule, but enough quality, individualized interventions should be planned to meet the identified outcomes for that patient.
Creating Nursing Care Plans
Nursing care plans are created by registered nurses (RNs). Documentation of individualized nursing care plans are legally required in long-term care facilities by the Centers for Medicare and Medicaid Services (CMS) and in hospitals by The Joint Commission. CMS guidelines state, “Residents and their representative(s) must be afforded the opportunity to participate in their care planning process and to be included in decisions and changes in care, treatment, and/or interventions. This applies both to initial decisions about care and treatment, as well as the refusal of care or treatment. Facility staff must support and encourage participation in the care planning process. This may include ensuring that residents, families, or representatives understand the comprehensive care planning process, holding care planning meetings at the time of day when a resident is functioning best and patient representatives can be present, providing sufficient notice in advance of the meeting, scheduling these meetings to accommodate a resident’s representative (such as conducting the meeting in-person, via a conference call, or video conferencing), and planning enough time for information exchange and decision-making. A resident has the right to select or refuse specific treatment options before the care plan is instituted.” [8] The Joint Commission conceptualizes the care planning process as the structuring framework for coordinating communication that will result in safe and effective care. [9]
Many facilities have established standardized nursing care plans with lists of possible interventions that can be customized for each specific patient. Other facilities require the nurse to develop each care plan independently. Whatever the format, nursing care plans should be individualized to meet the specific and unique needs of each patient. See Figure 4.13 [10] for an image of a standardized care plan.
Nursing care plans created in nursing school can also be in various formats such as concept maps or tables. Some are fun and creative, while others are more formal. See Figure 4.14 [11] for an image of a creative care plan created by a nursing student. Appendix B contains a template that can be used for creating nursing care plans.
- American Nurses Association. (2021). Nursing : Scope and standards of practice (4th ed.). American Nurses Association. ↵
- Herdman, T. H., & Kamitsuru, S. (Eds.). (2018). Nursing diagnoses: Definitions and classification, 2018-2020 . Thieme Publishers New York. ↵
- Butcher, H. K., Bulechek, G. M., Dochterman, J. M., & Wagner, C. M. (2018). Nursing interventions classifications (NIC) (7th ed.). Elsevier. ↵
- " 400845937-huge.jpg " by Flamingo Images is used under license from Shutterstock.com ↵
- Butcher, H. K., Bulechek, G. M., Dochterman, J. M., & Wagner, C. M. (2018). Nursing interventions classifications (NIC) (7th ed.). Elsevier. ↵
- NCSBN. (n.d.). 2019 NCLEX-RN test plan . https://www.ncsbn.org/2019_RN_TestPlan-English.htm ↵
- Vera, M. (2020). Nursing care plan (NCP): Ultimate guide and database. https://nurseslabs.com/nursing-care-plans/#:~:text=Collaborative%20interventions%20are%20actions%20that,to%20gain%20their%20professional%20viewpoint. ↵
- Centers for Medicare and Medicaid Services. (2017). State operations manual: Appendix PP - Guidance to surveyors for long term care facilities. https://www.cms.gov/Regulations-and-Guidance/Guidance/Manuals/downloads/som107ap_pp_guidelines_ltcf.pdf ↵
- The Joint Commission (n.d.). Standards and guides pertinent to nursing practice . https://www.jointcommission.org/resources/for-nurses/nursing-resources/ ↵
- " Figure 3-3. An example of a nursing care plan in an Australian residential aged care home..png " by NurseRecord is licensed under CC BY-SA 4.0 ↵
- “Creative Student Nurse Care Plan” by Janessa Lepak, RN for Chippewa Valley Technical College is licensed under CC BY 4.0 ↵


7.3: Introduction to Employee Compensation, Incentive, and Benefits Strategies
What you’ll learn to do: describe common employee compensation, incentive, and benefits strategies
Most people work to make a living. In other words, our jobs make it possible for us to buy a home, food, transportation, clothing, and the extras that make life fun. For most of us, the idea of making more rather than less money is compelling—and may draw us to a job or keep us in a job.
But for many people, money is just part of the larger picture. Some of us choose a particular job because we care about the products, services, or goals of the company. Or we need the generous health and education benefits provided to employees. Some people may choose a job because of its flexible hours, ample paid vacation, or opportunities to grow.
Employee compensation, incentives, and benefits can be combined in many different ways to suit the needs, desires, and challenges of each company and employee.
Contributors and Attributions
- Introduction to Employee Compensation, Incentive, and Benefits Strategies. Authored by : Lisa Jo Rudy and Lumen Learning. License : CC BY: Attribution


11.3: Aleatoric Music
Aleatoric music involves the use of chance in either the composition or performance of the piece. John Cage became a strong proponent of aleatoric techniques, even going so far as to use them in lectures as well as musical compositions. To be clear, the Sonatas and Interludes are not aleatoric works, so the John Cage piece on our playlist does not incorporate aleatoric elements. However, John Cage is so associated with chance music that it would seem odd not to study this important 20th century technique in connection with Cage.
Introduction
Aleatoric music (also aleatory music or chance music ; from the Latin word alea , meaning “dice”) is music in which some element of the composition is left to chance, and/or some primary element of a composed work’s realization is left to the determination of its performer(s). The term is most often associated with procedures in which the chance element involves a relatively limited number of possibilities.
Types of Indeterminate Music
Some writers do not make a distinction between aleatory, chance, and indeterminancy in music, and use the terms interchangeably. From this point of view, indeterminate or chance music can be divided into three groups: (1) the use of random procedures to produce a determinate, fixed score, (2) mobile form, and (3) indeterminate notation, including graphic notation and texts.
The first group includes scores in which the chance element is involved only in the process of composition, so that every parameter is fixed before their performance. In John Cage’s Music of Changes (1951), for example, the composer selected duration, tempo, and dynamics by using the I Ching , an ancient Chinese book which prescribes methods for arriving at random numbers. Because this work is absolutely fixed from performance to performance, Cage regarded it as an entirely determinate work made using chance procedures. On the level of detail, Iannis Xenakis used probability theories to define some microscopic aspects of Pithoprakta (1955–56), which is Greek for “actions by means of probability.” This work contains four sections, characterized by textural and timbral attributes, such as glissandi and pizzicati. At the macroscopic level, the sections are designed and controlled by the composer while the single components of sound are controlled by mathematical theories.
Contributors and Attributions
- Authored by : Elliott Jones. Provided by : Santa Ana College. Located at : http://www.sac.edu . License : CC BY: Attribution
- Aleatoric music. Provided by : Wikipedia. Located at : http://en.Wikipedia.org/wiki/Aleatoric_music . License : CC BY-SA: Attribution-ShareAlike


6.4: Graphs of Logarithmic Functions
- Identify the domain of a logarithmic function.
- Graph logarithmic functions.
In the Section on Graphs of Exponential Functions , we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the cause for an effect .
To illustrate, suppose we invest \($2500\) in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year \(t\) can be found with the equation \(A=2500e^{0.05t}\).
But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure \(\PageIndex{1}\) shows this point on the logarithmic graph.
In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions.
Finding the Domain of a Logarithmic Function
Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.
Recall that the exponential function is defined as \(y=b^x\) for any real number \(x\) and constant \(b>0\), \(b≠1\), where
- The domain of \(y\) is \((−\infty,\infty)\).
- The range of \(y\) is \((0,\infty)\).
In the last section we learned that the logarithmic function \(y={\log}_b(x)\) is the inverse of the exponential function \(y=b^x\). So, as inverse functions:
- The domain of \(y={\log}_b(x)\) is the range of \(y=b^x\): \((0,\infty)\).
- The range of \(y={\log}_b(x)\) is the domain of \(y=b^x\): \((−\infty,\infty)\).
Transformations of the parent function \(y={\log}_b(x)\) behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, stretches, compressions, and reflections—to the parent function without loss of shape.
In Graphs of Exponential Functions we saw that certain transformations can change the range of \(y=b^x\). Similarly, applying transformations to the parent function \(y={\log}_b(x)\) can change the domain . When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers . That is, the argument of the logarithmic function must be greater than zero.
For example, consider \(f(x)={\log}_4(2x−3)\). This function is defined for any values of \(x\) such that the argument, in this case \(2x−3\),is greater than zero. To find the domain, we set up an inequality and solve for \(x\):
\[\begin{align*} 2x-3&> 0 \qquad \text {Show the argument greater than zero}\\ 2x&> 3 \qquad \text{Add 3} \\ x&> 1.5 \qquad \text{Divide by 2} \\ \end{align*}\]
In interval notation, the domain of \(f(x)={\log}_4(2x−3)\) is \((1.5,\infty)\).
- Set up an inequality showing the argument greater than zero.
- Solve for \(x\).
- Write the domain in interval notation.
What is the domain of \(f(x)={\log}_2(x+3)\)?
Solution
The logarithmic function is defined only when the input is positive, so this function is defined when \(x+3>0\). Solving this inequality,
\[\begin{align*} x+3&> 0 \qquad \text{The input must be positive}\\ x&> -3 \qquad \text{Subtract 3} \end{align*}\]
The domain of \(f(x)={\log}_2(x+3)\) is \((−3,\infty)\).
What is the domain of \(f(x)={\log}_5(x−2)+1\)?
- Answer
-
\((2,\infty)\)
What is the domain of \(f(x)=\log(5−2x)\)?
Solution
The logarithmic function is defined only when the input is positive, so this function is defined when \(5–2x>0\). Solving this inequality,
\[\begin{align*} 5-2x&> 0 \qquad \text{The input must be positive}\\ -2x&> -5 \qquad \text{Subtract 5}\\ x&< \dfrac{5}{2} \qquad \text{Divide by -2 and switch the inequality} \end{align*}\]
The domain of \(f(x)=\log(5−2x)\) is \(\left(–\infty,\dfrac{5}{2}\right)\).
What is the domain of \(f(x)=\log(x−5)+2\)?
- Answer
-
\((5,\infty)\)
Graphing Logarithmic Functions
Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function \(y={\log}_b(x)\) along with all its transformations: shifts, stretches, compressions, and reflections.
We begin with the parent function \(y={\log}_b(x)\). Because every logarithmic function of this form is the inverse of an exponential function with the form \(y=b^x\), their graphs will be reflections of each other across the line \(y=x\). To illustrate this, we can observe the relationship between the input and output values of \(y=2^x\) and its equivalent \(x={\log}_2(y)\) in Table \(\PageIndex{1}\).
| \(x\) | \(−3\) | \(−2\) | \(−1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|---|---|
| \(2^x=y\) | \(\dfrac{1}{8}\) | \(\dfrac{1}{4}\) | \(\dfrac{1}{2}\) | \(1\) | \(2\) | \(4\) | \(8\) |
| \({\log}_2(y)=x\) | \(−3\) | \(−2\) | \(−1\) | \(0\) | \(1\) | \(2\) | \(3\) |
Using the inputs and outputs from Table \(\PageIndex{1}\), we can build another table to observe the relationship between points on the graphs of the inverse functions \(f(x)=2^x\) and \(g(x)={\log}_2(x)\). See Table \(\PageIndex{2}\).
| \(f(x)=2^x\) | \(\left(−3,\dfrac{1}{8}\right)\) | \(\left(−2,\dfrac{1}{4}\right)\) | \(\left(−1,\dfrac{1}{2}\right)\) | \((0,1)\) | \((1,2)\) | \((2,4)\) | \((3,8)\) |
|---|---|---|---|---|---|---|---|
| \(g(x)={\log}_2(x)\) | \(\left(\dfrac{1}{8},−3\right)\) | \(\left(\dfrac{1}{4},−2\right)\) | \(\left(\dfrac{1}{2},−1\right)\) | \((1,0)\) | \((2,1)\) | \((4,2)\) | \((8,3)\) |
As we’d expect, the \(x\)- and \(y\)-coordinates are reversed for the inverse functions. Figure \(\PageIndex{2}\) shows the graph of \(f\) and \(g\).
Observe the following from the graph:
- \(f(x)=2^x\) has a \(y\)-intercept at \((0,1)\) and \(g(x)={\log}_2(x)\) has an \(x\)- intercept at \((1,0)\).
- The domain of \(f(x)=2^x\), \((−\infty,\infty)\), is the same as the range of \(g(x)={\log}_2(x)\).
- The range of \(f(x)=2^x\), \((0,\infty)\), is the same as the domain of \(g(x)={\log}_2(x)\).
For any real number \(x\) and constant \(b>0\), \(b≠1\), we can see the following characteristics in the graph of \(f(x)={\log}_b(x)\):
- one-to-one function
- vertical asymptote: \(x=0\)
- domain: \((0,\infty)\)
- range: \((−\infty,\infty)\)
- \(x\) - intercept: \((1,0)\) and key point \((b,1)\)
- \(y\)-intercept: none
- increasing if \(b>1\)
- decreasing if \(0<b<1\)
See Figure \(\PageIndex{3}\).
Figure \(\PageIndex{4}\) shows how changing the base \(b\) in \(f(x)={\log}_b(x)\) can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (Note: recall that the function \(\ln(x)\) has base \(e≈2.718\).)
- Draw and label the vertical asymptote, \(x=0\).
- Plot the x- intercept, \((1,0)\).
- Plot the key point \((b,1)\).
- Draw a smooth curve through the points.
- State the domain, \((0,\infty)\),the range, \((−\infty,\infty)\), and the vertical asymptote, \(x=0\).
Graph \(f(x)={\log}_5(x)\). State the domain, range, and asymptote.
Solution
Before graphing, identify the behavior and key points for the graph.
- Since \(b=5\) is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote \(x=0\), and the right tail will increase slowly without bound.
- The \(x\)-intercept is \((1,0)\).
- The key point \((5,1)\) is on the graph.
- We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points (see Figure \(\PageIndex{5}\)).
The domain is \((0,\infty)\), the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=0\).
Graph \(f(x)={\log}_{\tfrac{1}{5}}(x)\). State the domain, range, and asymptote.
- Answer
-
The domain is \((0,\infty)\), the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=0\).
Graphing Transformations of Logarithmic Functions
As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the parent function \(y={\log}_b(x)\) without loss of shape.
Graphing a Horizontal Shift of \(f(x) = log_b(x)\)
When a constant \(c\) is added to the input of the parent function \(f(x)={\log}_b(x)\), the result is a horizontal shift \(c\) units in the opposite direction of the sign on \(c\). To visualize horizontal shifts, we can observe the general graph of the parent function \(f(x)={\log}_b(x)\) and for \(c>0\) alongside the shift left, \(g(x)={\log}_b(x+c)\), and the shift right, \(h(x)={\log}_b(x−c)\). See Figure \(\PageIndex{7}\).
For any constant \(c\),the function \(f(x)={\log}_b(x+c)\)
- shifts the parent function \(y={\log}_b(x)\) left \(c\) units if \(c>0\).
- shifts the parent function \(y={\log}_b(x)\) right \(c\) units if \(c<0\).
- has the vertical asymptote \(x=−c\).
- has domain \((−c,\infty)\).
- has range \((−\infty,\infty)\).
-
Identify the horizontal shift:
- If \(c>0\),shift the graph of \(f(x)={\log}_b(x)\) left \(c\) units.
- If \(c<0\),shift the graph of \(f(x)={\log}_b(x)\) right \(c\) units.
- Draw the vertical asymptote \(x=−c\).
- Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting \(c\) from the \(x\) coordinate.
- Label the three points.
- The Domain is \((−c,\infty)\),the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=−c\).
Sketch the horizontal shift \(f(x)={\log}_3(x−2)\) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.
Solution
Since the function is \(f(x)={\log}_3(x−2)\), we notice \(x+(−2)=x–2\).
Thus \(c=−2\), so \(c<0\). This means we will shift the function \(f(x)={\log}_3(x)\) right 2 units.
The vertical asymptote is \(x=−(−2)\) or \(x=2\).
Consider the three key points from the parent function, \(\left(\dfrac{1}{3},−1\right)\), \((1,0)\),and \((3,1)\).
The new coordinates are found by adding \(2\) to the \(x\) coordinates.
Label the points \(\left(\dfrac{7}{3},−1\right)\), \((3,0)\),and \((5,1)\).
The domain is \((2,\infty)\),the range is \((−\infty,\infty)\),and the vertical asymptote is \(x=2\).
Sketch a graph of \(f(x)={\log}_3(x+4)\) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.
- Answer
-
The domain is \((−4,\infty)\),the range \((−\infty,\infty)\),and the asymptote \(x=–4\).
Graphing a Vertical Shift of \(y = log_b(x)\)
When a constant \(d\) is added to the parent function \(f(x)={\log}_b(x)\),the result is a vertical shift \(d\) units in the direction of the sign on \(d\). To visualize vertical shifts, we can observe the general graph of the parent function \(f(x)={\log}_b(x)\) alongside the shift up, \(g(x)={\log}_b(x)+d\) and the shift down, \(h(x)={\log}_b(x)−d\).See Figure \(\PageIndex{10}\).
For any constant \(d\),the function \(f(x)={\log}_b(x)+d\)
- shifts the parent function \(y={\log}_b(x)\) up \(d\) units if \(d>0\).
- shifts the parent function \(y={\log}_b(x)\) down \(d\) units if \(d<0\).
- has the vertical asymptote \(x=0\).
- has domain \((0,\infty)\).
- has range \((−\infty,\infty)\).
-
Identify the vertical shift:
- If \(d>0\), shift the graph of \(f(x)={\log}_b(x)\) up \(d\) units.
- If \(d<0\), shift the graph of \(f(x)={\log}_b(x)\) down \(d\) units.
- Draw the vertical asymptote \(x=0\).
- Identify three key points from the parent function. Find new coordinates for the shifted functions by adding \(d\) to the \(y\) coordinate.
- Label the three points.
- The domain is \((0,\infty)\), the range is \((−\infty,\infty)\),and the vertical asymptote is \(x=0\).
Sketch a graph of \(f(x)={\log}_3(x)−2\) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Solution
Since the function is \(f(x)={\log}_3(x)−2\),we will notice \(d=–2\). Thus \(d<0\).
This means we will shift the function \(f(x)={\log}_3(x)\) down \(2\) units.
The vertical asymptote is \(x=0\).
Consider the three key points from the parent function, \(\left(\dfrac{1}{3},−1\right)\), \((1,0)\),and \((3,1)\).
The new coordinates are found by subtracting \(2\) from the y coordinates.
Label the points \(\left(\dfrac{1}{3},−3\right)\), \((1,−2)\), and \((3,−1)\).
The domain is \((0,\infty)\),the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=0\).
The domain is \((0,\infty)\),the range is \((−\infty,\infty)\),and the vertical asymptote is \(x=0\).
Sketch a graph of \(f(x)={\log}_2(x)+2\) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
- Answer
-
The domain is \((0,\infty)\),the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=0\).
Graphing Stretches and Compressions of \(y = log_b(x)\)
When the parent function \(f(x)={\log}_b(x)\) is multiplied by a constant \(a>0\), the result is a vertical stretch or compression of the original graph. To visualize stretches and compressions, we set \(a>1\) and observe the general graph of the parent function \(f(x)={\log}_b(x)\) alongside the vertical stretch, \(g(x)=a{\log}_b(x)\) and the vertical compression, \(h(x)=\dfrac{1}{a}{\log}_b(x)\).See Figure \(\PageIndex{13}\).
For any constant \(a>1\),the function \(f(x)=a{\log}_b(x)\)
- stretches the parent function \(y={\log}_b(x)\) vertically by a factor of \(a\) if \(a>1\).
- compresses the parent function \(y={\log}_b(x)\) vertically by a factor of \(a\) if \(0<a<1\).
- has the vertical asymptote \(x=0\).
- has the \(x\)-intercept \((1,0)\).
- has domain \((0,\infty)\).
- has range \((−\infty,\infty)\).
-
Identify the vertical stretch or compressions:
- If \(|a|>1\), the graph of \(f(x)={\log}_b(x)\) is stretched by a factor of \(a\) units.
- If \(|a|<1\), the graph of \(f(x)={\log}_b(x)\) is compressed by a factor of \(a\) units.
- Draw the vertical asymptote \(x=0\).
- Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying they y coordinates by \(a\).
- Label the three points.
- The domain is \((0,\infty)\), the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=0\).
Sketch a graph of \(f(x)=2{\log}_4(x)\) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Solution
Since the function is \(f(x)=2{\log}_4(x)\),we will notice \(a=2\).
This means we will stretch the function \(f(x)={\log}_4(x)\) by a factor of \(2\).
The vertical asymptote is \(x=0\).
Consider the three key points from the parent function, \(\left(\dfrac{1}{4},−1\right)\), \((1,0)\), and \((4,1)\).
The new coordinates are found by multiplying the \(y\) coordinates by \(2\).
Label the points \(\left(\dfrac{1}{4},−2\right)\), \((1,0)\), and \((4,2)\).
The domain is \((0, \infty)\), the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=0\). See Figure \(\PageIndex{14}\).
The domain is \((0,\infty)\), the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=0\).
Sketch a graph of \(f(x)=\dfrac{1}{2}{\log}_4(x)\) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
- Answer
-
The domain is \((0,\infty)\), the range is \((−\infty,\infty)\),and the vertical asymptote is \(x=0\).
Sketch a graph of \(f(x)=5{\log}(x+2)\). State the domain, range, and asymptote.
Solution
Remember: what happens inside parentheses happens first. First, we move the graph left \(2\) units, then stretch the function vertically by a factor of \(5\), as in Figure \(\PageIndex{16}\). The vertical asymptote will be shifted to \(x=−2\). The \(x\)-intercept will be \((−1,0)\). The domain will be \((−2,\infty)\). Two points will help give the shape of the graph: \((−1,0)\) and \((8,5)\). We chose \(x=8\) as the x -coordinate of one point to graph because when \(x=8\), \(x+2=10\), the base of the common logarithm.
The domain is \((−2,\infty)\), the range is \((−\infty,\infty)\),and the vertical asymptote is \(x=−2\).
Sketch a graph of the function \(f(x)=3{\log}(x−2)+1\). State the domain, range, and asymptote.
- Answer
-
The domain is \((2,\infty)\),the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=2\).
Graphing Reflections of \(f(x) = log_b(x)\)
When the parent function \(f(x)={\log}_b(x)\) is multiplied by \(−1\),the result is a reflection about the \(x\)-axis. When the input is multiplied by \(−1\),the result is a reflection about the \(y\)-axis. To visualize reflections, we restrict \(b>1\), and observe the general graph of the parent function \(f(x)={\log}_b(x)\) alongside the reflection about the \(x\)-axis, \(g(x)=−{\log}_b(x)\) and the reflection about the \(y\)-axis, \(h(x)={\log}_b(−x)\).
The function \(f(x)=−{\log}_b(x)\)
- reflects the parent function \(y={\log}_b(x)\) about the \(x\)-axis.
- has domain, \((0,\infty)\), range, \((−\infty,\infty)\), and vertical asymptote, \(x=0\), which are unchanged from the parent function.
The function \(f(x)={\log}_b(−x)\)
- reflects the parent function \(y={\log}_b(x)\) about the \(y\)-axis.
- has domain \((−\infty,0)\).
- has range, \((−\infty,\infty)\), and vertical asymptote, \(x=0\), which are unchanged from the parent function.
| If \(f(x)=−{\log}_b(x)\) | If \(f(x)={\log}_b(−x)\) |
|---|---|
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Sketch a graph of \(f(x)=\log(−x)\) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Solution
Before graphing \(f(x)=log(−x)\), \(f(x)=log(−x)\),identify the behavior and key points for the graph.
- Since \(b=10\) is greater than one, we know that the parent function is increasing. Since the input value is multiplied by \(−1\), \(f(x)\) is a reflection of the parent graph about the \(y\) - axis. Thus, \(f(x)=\log(−x)\) will be decreasing as \(x\) moves from negative infinity to zero, and the right tail of the graph will approach the vertical asymptote \(x=0\).
- The \(x\)-intercept is \((−1,0)\).
- We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points.
The domain is \((−\infty,0)\), the range is \((−\infty,\infty)\), and the vertical asymptote is \(x=0\).
Graph \(f(x)=−\log(−x)\). State the domain, range, and asymptote.
- Answer
-
The domain is \((−\infty,0)\),the range is \((−\infty,\infty)\),and the vertical asymptote is \(x=0\).
- Press [Y=] . Enter the given logarithm equation or equations as Y 1 = and, if needed, Y 2 = .
- Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection.
- To find the value of \(x\), we compute the point of intersection. Press [2ND] then [CALC] . Select “intersect” and press [ENTER] three times. The point of intersection gives the value of \(x\),for the point(s) of intersection.
Solve \(4\ln(x)+1=−2\ln(x−1)\) graphically. Round to the nearest thousandth.
Solution
Press [Y=] and enter \(4\ln(x)+1\) next to Y 1 =. Then enter \(−2\ln(x−1)\) next to Y 2 = . For a window, use the values \(0\) to \(5\) for \(x\0 and \(–10\) to \(10\) for \(y\). Press [GRAPH] . The graphs should intersect somewhere a little to right of \(x=1\).
For a better approximation, press [2ND] then [CALC] . Select [5: intersect] and press [ENTER] three times. The x -coordinate of the point of intersection is displayed as \(1.3385297\). (Your answer may be different if you use a different window or use a different value for Guess? ) So, to the nearest thousandth, \(x≈1.339\).
Solve \(5\log(x+2)=4−\log(x)\) graphically. Round to the nearest thousandth.
- Answer
-
\(x≈3.049\)
Summarizing Translations of the Logarithmic Function
Now that we have worked with each type of translation for the logarithmic function, we can summarize each in Table \(\PageIndex{4}\) to arrive at the general equation for translating exponential functions.
| Translations of the Parent Function \(y={\log}_b(x)\) | |
|---|---|
| Translation | Form |
|
Shift Horizontally \(c\) units to the left Vertically \(d\) units up |
\(y={\log}_b(x+c)+d\) |
|
Stretch and Compress Stretch if \(|a|>1\) Compression if \(|a|<1\) |
\(y=a{\log}_b(x)\) |
| Reflect about the x-axis | \(y=−{\log}_b(x)\) |
| Reflect about the y-axis | \(y={\log}_b(−x)\) |
| General equation for all translations | \(y=a{\log}_b(x+c)+d\) |
All translations of the parent logarithmic function, \(y={\log}_b(x)\), have the form
\(f(x)=a{\log}_b(x+c)+d\)
where the parent function, \(y={\log}_b(x)\), \(b>1\),is
- shifted vertically up \(d\) units.
- shifted horizontally to the left \(c\) units.
- stretched vertically by a factor of \(|a|\) if \(|a|>0\).
- compressed vertically by a factor of \(|a|\) if \(0<|a|<1\).
- reflected about the x- axis when \(a<0\).
For \(f(x)=\log(−x)\), the graph of the parent function is reflected about the y -axis.
What is the vertical asymptote of \(f(x)=−2{\log}_3(x+4)+5\)?
Solution
The vertical asymptote is at \(x=−4\).
Analysis
The coefficient, the base, and the upward translation do not affect the asymptote. The shift of the curve \(4\) units to the left shifts the vertical asymptote to\(x=−4\).
What is the vertical asymptote of \(f(x)=3+\ln(x−1)\)?
- Answer
-
\(x=1\)
Find a possible equation for the common logarithmic function graphed in Figure \(\PageIndex{21}\).
Solution
This graph has a vertical asymptote at \(x=–2\) and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form:
\(f(x)=−a\log(x+2)+k\)
It appears the graph passes through the points \((–1,1)\) and \((2,–1)\). Substituting \((–1,1)\),
\[\begin{align*} 1&= -a\log(-1+2)+k \qquad \text{Substitute} (-1,1)\\ 1&= -a\log(1)+k \qquad \text{Arithmetic}\\ 1&= k\log(1)\\ &= 0 \end{align*}\]
Next, substituting in \((2,–1)\),
\[\begin{align*} -1&= -a\log(2+2)+1 \qquad \text{Substitute} (2,-1)\\ -2&= -a\log(4) \qquad \text{Arithmetic}\\ a&= \dfrac{2}{\log(4)} \qquad \text{Solve for a} \end{align*}\]
This gives us the equation \(f(x)=–\dfrac{2}{\log(4)}\log(x+2)+1\).
Analysis
We can verify this answer by comparing the function values in Table \(\PageIndex{5}\) with the points on the graph in Figure \(\PageIndex{21}\).
| \(x\) | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|
| \(f(x)\) | 1 | 0 | −0.58496 | −1 | −1.3219 |
| \(x\) | 4 | 5 | 6 | 7 | 8 |
| \(f(x)\) | −1.5850 | −1.8074 | −2 | −2.1699 | −2.3219 |
Give the equation of the natural logarithm graphed in Figure \(\PageIndex{22}\).
- Answer
-
\(f(x)=2\ln(x+3)−1\)
Yes, if we know the function is a general logarithmic function. For example, look at the graph in Figure \(\PageIndex{22}\). The graph approaches \(x=−3\) (or thereabouts) more and more closely, so \(x=−3\) is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, \({x|x>−3}\). The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as \(x\rightarrow −3^+\), \(f(x)\rightarrow −\infty\) and as \(x\rightarrow \infty\), \(f(x)\rightarrow \infty\).
Access these online resources for additional instruction and practice with graphing logarithms.
- Graph an Exponential Function and Logarithmic Function
- Match Graphs with Exponential and Logarithmic Functions
- Find the Domain of Logarithmic Functions
Key Equations
| General Form for the Translation of the Parent Logarithmic Function \(f(x)={\log}_b(x)\) | \(f(x)=a{\log}_b(x+c)+d\) |
Key Concepts
- To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for \(x\). See Example \(\PageIndex{1}\) and Example \(\PageIndex{2}\)
- The graph of the parent function \(f(x)={\log}_b(x)\) has an x- intercept at \((1,0)\),domain \((0,\infty)\),range \((−\infty,\infty)\),vertical asymptote \(x=0\),and
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- if \(b>1\),the function is increasing.
- if \(0<b<1\), the function is decreasing.
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The equation \(f(x)={\log}_b(x+c)\) shifts the parent function \(y={\log}_b(x)\) horizontally
- left \(c\) units if \(c>0\).
- right \(c\) units if \(c<0\).
-
The equation \(f(x)={\log}_b(x)+d\) shifts the parent function \(y={\log}_b(x)\) vertically
- up \(d\) units if \(d>0\).
- down \(d\) units if \(d<0\).
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For any constant \(a>0\), the equation \(f(x)=a{\log}_b(x)\)
- stretches the parent function \(y={\log}_b(x)\) vertically by a factor of \(a\) if \(|a|>1\).
- compresses the parent function \(y={\log}_b(x)\) vertically by a factor of \(a\) if \(|a|<\)1.
-
When the parent function \(y={\log}_b(x)\) is multiplied by \(−1\), the result is a reflection about the
x
-axis. When the input is multiplied by \(−1\), the result is a reflection about the
y
-axis.
- The equation \(f(x)=−{\log}_b(x)\) represents a reflection of the parent function about the x- axis.
- The equation \(f(x)={\log}_b(−x)\) represents a reflection of the parent function about the y- axis.
- A graphing calculator may be used to approximate solutions to some logarithmic equations See Example \(\PageIndex{9}\).
- All translations of the logarithmic function can be summarized by the general equation \(f(x)=a{\log}_b(x+c)+d\). See Table \(\PageIndex{4}\).
- Given an equation with the general form \(f(x)=a{\log}_b(x+c)+d\),we can identify the vertical asymptote \(x=−c\) for the transformation. See Example \(\PageIndex{10}\).
- Using the general equation \(f(x)=a{\log}_b(x+c)+d\), we can write the equation of a logarithmic function given its graph. See Example \(\PageIndex{11}\).


25.0: Introduction
The lower respiratory system (shown in Figure 25.2) plays a crucial role in respiration, allowing the exchange of oxygen and carbon dioxide between the body and the external environment. The lower respiratory system includes the trachea, the lungs, and within the lungs, the bronchi, bronchioles, and alveoli. See Introduction to the Respiratory System for more information on the lower and upper respiratory systems. Various respiratory diseases and disorders can significantly affect the functioning of the lower respiratory system, leading to respiratory distress, impaired lung function, and reduced quality of life. Lower respiratory drugs are a class of medications designed to alleviate symptoms, manage chronic conditions, and improve overall respiratory function.
Respiratory diseases affecting the lower respiratory system can be broadly classified into two categories: obstructive and restrictive lung diseases. Obstructive lung diseases, such as asthma and Chronic obstructive pulmonary disease (COPD) , are characterized by airflow limitation due to inflammation, bronchoconstriction, and increased mucus production. Asthma affects the lungs, causing repeated episodes of wheezing, breathlessness, and coughing. COPD is a group of diseases including emphysema and chronic bronchitis. Emphysema damages the airways and causes chronic inflammation in the lungs. This makes it very difficult for the person to breathe. Chronic bronchitis is a productive cough of more than 3 months, specifically within a span of 2 years (Widysanto & Mathew, 2022).
Restrictive lung diseases, such as pulmonary fibrosis , involve a reduced lung capacity due to impaired lung tissue expansion or chest wall abnormalities (Martinez-Pitre et al., 2023).
Lower respiratory drugs encompass a wide range of pharmacological agents, each targeting specific mechanisms involved in respiratory diseases. These drugs can be administered via various routes, including inhalation, oral ingestion, and intravenous infusion, depending on the medication’s characteristics and the severity of the respiratory condition.


4.2: The Role of Socialization
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- 4.2.1: The Role of Socialization
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- <IP_ADDRESS>: Childhood Socialization
- <IP_ADDRESS>A: Child Socialization
- <IP_ADDRESS>B: Theoretical Perspectives on Childhood Socialization
- <IP_ADDRESS>C: Identity Formation
- 4.2.1A: The Role of Socialization
- 4.2.1B: Nature vs. Nurture- A False Debate
- 4.2.1C: Sociobiology
- 4.2.1D: Deprivation and Development
- 4.2.1E: Isolation and Development
- 4.2.1F: Feral Children
- 4.2.1G: Institutionalized Children
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- 4.2.5: Agents of Socialization
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- 4.2.5A: Family
- 4.2.5B: Neighborhood
- 4.2.5C: School
- 4.2.5D: Day Care
- 4.2.5E: Peer Groups
- 4.2.5F: Mass Media and Technology
- 4.2.5G: Religion
- 4.2.5H: Workplace
- 4.2.5I: The Division of Labor
- 4.2.5J: The Incest Taboo, Marriage, and the Family
- 4.2.5K: Ideology
- 4.2.5L: Resocialization and Total Institutions