Skip to main content
Study Questions
-
What is the difference between data, information, and knowledge?
-
Explain in your own words the difference between hardware and software components of information systems.
-
What is the difference between quantitative data and qualitative data? In what situations could the number 63 be considered qualitative data?
-
What are the characteristics of a relational database?
-
When would using a personal DBMS make sense?
-
What is the difference between a spreadsheet and a database? List three differences between them.
-
Describe what the term normalization means.
-
What is Big Data?
-
Name a database you interact with frequently. What would some of the field names be?
-
Describe the benefits and what open-source data is.
-
Name three advantages of using a data warehouse.
-
What is data mining?
Exercises
-
Review the design of the Student Clubs database earlier in this chapter. Reviewing the lists of data types given, what data types would you assign to each of the fields in each of the tables. What lengths would you assign to the text fields?
-
Review structured and unstructured data and list five reasons to use each.
-
Using Microsoft Access, download the database file of comprehensive baseball statistics from the website
-
SeanLahman.com
. (If you don’t have Microsoft Access, you can download an abridged version of the file
here
that is compatible with Apache Open Office). Review the structure of the tables included in the database. Come up with three different data-mining experiments you would like to try, and explain which fields in which tables would have to be analyzed.
-
Do some original research and find two examples of data mining. Summarize each example and then write about what the two examples have in common.
-
Conduct some independent research on the process of business intelligence. Using at least two scholarly or practitioner sources, write a two-page paper giving examples of how business intelligence is being used.
-
Conduct some independent research on the latest technologies being used for knowledge management. Using at least two scholarly or practitioner sources, write a two-page paper giving examples of software applications or new technologies being used in this field.


6.1: Introduction to Permutation Groups
Definition: Permutation
A permutation on a set \(A\) is a bijection from \(A\) to \(A\text{.}\) We say a permutation \(\sigma\) on \(A\) fixes \(a\in A\) if \(\sigma(a)=a\text{.}\)
Example \(\PageIndex{1}\)
Let \(A\) be the set \(A=\{\Delta, \star, 4\}\text{.}\) Then the functions \(\sigma : A\to A\) defined by
\begin{equation*} \sigma(\Delta)=\star, \sigma(\star)=\Delta, \text{and } \sigma(4)=4; \end{equation*}
and \(\tau : A\to A\) defined by
\begin{equation*} \tau(\Delta)=4, \tau(\star)=\Delta, \text{and } \tau(4)=\star \end{equation*}
are both permutations on \(A\text{.}\)
Definition: Permutation Multiplication
Composition of permutations on a set \(A\) is often called permutation multiplication , and if \(\sigma\) and \(\tau\) are permutations on a set \(A\text{,}\) we usually omit the composition symbol and write \(\sigma \circ \tau\) simply as \(\sigma \tau\text{.}\)
Note
For us, if \(\sigma\) and \(\tau\) are permutations on a set \(A\text{,}\) then applying \(\sigma \tau\) to \(A\) means first applying \(\tau\) and then applying \(\sigma\text{.}\) This is due to the conventional right-to-left reading of function compositions.
That is, if \(a\in A\text{,}\) by \(\sigma \tau(a)\) we mean \(\sigma(\tau(a))\text{.}\) (Some other books/mathematicians do not use this convention, and read permutation multiplication from left-to-right. Make sure to always know what convention your particular author or colleague is using!)
Example \(\PageIndex{2}\)
Let \(A\text{,}\) \(\sigma\text{,}\) and \(\tau\) be as in Example \(6.1.1\). Then \(\sigma \tau\) is the function from \(A\) to \(A\) defined by
\begin{equation*} \sigma \tau(\Delta)=4, \sigma \tau(\star)=\star, \text{and } \sigma \tau(4)=\Delta, \end{equation*}
while \(\tau \sigma\) is the function from \(A\) to \(A\) defined by
\begin{equation*} \tau \sigma (\Delta)=\Delta, \tau \sigma (\star)=4, \text{and } \tau \sigma(4)=\star \end{equation*}
Definition
Given a set \(A\text{,}\) we define \(S_A\) to be the set of all permutations on \(A\text{.}\)
Theorem \(\PageIndex{1}\)
Given a set \(A\text{:}\)
-
\(S_A\) is a group under permutation multiplication.
-
If \(A\) has finite cardinality \(n\text{,}\) then \(|S_A|=n!\) (if \(|A|=\infty\) then \(|S_A|\) is also \(\infty\)).
-
\(S_A\) is abelian if \(|A|=1\) or \(2\text{,}\) and nonabelian otherwise.
- Proof
-
1. Let \(σ,τ∈S_A\). Since a composition of bijections is a bijection (see Theorem \(1.2.3\)), \(στ\) is a bijection from \(A\) to \(A\), hence is in \(S_A\). So \(\langle S_A, \circ \rangle\) is a binary structure
Next, function composition is always associative.
The identity function \(1_A:A→A\) defined by
\(1_A(a)=a \text{ for all } a∈A\)
clearly acts as an identity element in \(S_A\). Henceforth, we will denote \(1_A\) by \(e\).
Finally, let \(σ∈S_A\). Since \(σ\) is a bijection, \(σ\) has an inverse function \(σ^{−1}\) that is also a bijection from \(A\) to \(A\) (Theorem \(1.2.1\)). Since \(σ^{−1}∈S_A\) with \(σσ^{−1}=σ^{−1}σ=1_A\), every element of \(S_A\) has an inverse element in \(S_A\).
So \(S_A\) is a group.
2. Clearly \(|S_A|=∞\) when \(|A|=∞\), and a straightforward combinatorial argument yields that when \(|A|=n∈\mathbb{Z}^+\), we have \(|S_A|=n!\).
3. Finally, if \(|A|=1\) or \(2\), then \(|S_A|=1!=1\) or \(|S_A|=2!=2\) so \(S_A\) must be abelian (as it's a group of order \(1\) or \(2\)). On the other hand, suppose \(|A|>2\). Then \(A\) contains at least three distinct elements, say \(x\), \(y\), and \(z\). Let \(σ\) be the permutation of \(A\) swapping \(x\) and \(y\) and fixing every other element of \(A\), and let \(τ\) be the permutation of \(A\) swapping \(y\) and \(z\) and fixing every other element of \(A\). Then \(στ(x)=y\) while \(τσ(x)=z\), so \(στ≠τσ\), and hence \(S_A\) is nonabelian.
We will in the future use language provided by the following definition:
Definition: Permutation Group
A group is said to be a permutation group if it is a subgroup of \(S_A\) for some set \(A\text{.}\)
Remark
Notice that if \(A\) and \(B\) are sets, then \(|A|=|B|\) if and only if \(S_A\simeq S_B\text{.}\)
Thus, for any set \(B\) with \(|B|=n \in \mathbb{Z}^+\text{,}\) we have \(S_B\simeq S_A\text{,}\) where \(A=\{1,2,\ldots,n\}\text{.}\) Since we are concerned in this course primarily with group structures which are invariant under isomorphism, we may focus now on groups of permutations on the set \(\{1,2,\ldots, n\}\) (\(n\in \mathbb{Z}^+\)).


2: Fitting Statistical Models to Data
This text is about constructing and testing mathematical models of evolution. In my view the best comparative approaches have two features. First, the most useful methods emphasize parameter estimation over test statistics and P-values. Ideal methods fit models that we care about and estimate parameters that have a clear biological interpretation. To be useful, methods must also recognize and quantify uncertainty in our parameter estimates. Second, many useful methods involve model selection, the process of using data to objectively select the best model from a set of possibilities. When we use a model selection approach, we take advantage of the fact that patterns in empirical data sets will reject some models as implausible and support the predictions of others. This sort of approach can be a nice way to connect the results of a statistical analysis to a particular biological question.
-
- 2.1: Introduction
- Evolution is the product of a thousand stories.  The net result of these individual life stories over broad spans of time is evolution. At first glance, it might seem impossible to model this process over more than one or two generations. And yet scientific progress relies on creating simple models and confronting them with data. How can we evaluate models that consider evolution over millions of generations?
-
- 2.2: Standard Statistical Hypothesis Testing
- Standard hypothesis testing approaches focus almost entirely on rejecting null hypotheses. In the framework (usually referred to as the frequentist approach to statistics) one first defines a null hypothesis. This null hypothesis represents your expectation if some pattern, such as a difference among groups, is not present, or if some process of interest were not occurring.
-
- 2.3: Maximum Likelihood
- Likelihood is defined as the probability, given a model and a set of parameter values, of obtaining a particular set of data. That is, given a mathematical description of the world, what is the probability that we would see the actual data that we have collected? To calculate a likelihood, we have to consider a particular model that may have generated the data. That model will almost always have parameter values that need to be specified. We can refer to this specified model as a hypothesis, H.
-
- 2.4: Bayesian Statistics
- Recent years have seen tremendous growth of Bayesian approaches in reconstructing phylogenetic trees and estimating their branch lengths. Although there are currently only a few Bayesian comparative methods, their number will certainly grow as comparative biologists try to solve more complex problems.
-
- 2.5: AIC versus Bayes
- When you compare Bayes factors, you assume that one of the models you are considering is actually the true model that generated your data, and calculate posterior probabilities based on that assumption. By contrast, AIC assumes that reality is more complex than any of your models, and you are trying to identify the model that most efficiently captures the information in your data. So even though both techniques are carrying out model selection, the basic philosophy of how these models differ.
-
- 2.6: Models and Comparative Methods
- One theme in the book is that I emphasize fitting models to data and estimating parameters. I think that this approach is very useful for the future of the field of comparative statistics for three main reasons. First, it is flexible; one can easily compare a wide range of competing models to your data. Second, it is extendable; one can create new models and automatically fit them into a preexisting framework for data analysis. Finally, it is powerful.


2.5: The Propaganda Model
selected template will load here
This action is not available.
Following Smythe’s exposition of the audience and Jurgen Habermas’s (1991) elucidation of mass media as providing an apparatus whereby elite sectors of society can transform the democratising potential of the public sphere, a series of leftist academic media scholars have attempted to delineate the precise methods by which the mass media operates as a distorting lens which represents the vested interests of economic elites.
Most prominent within this PE-centred approach has been the ‘propaganda model’ (PM) of mass media presented by Noam Chomsky and Edward Herman in Manufacturing Consent: The Political Economy of Mass Media (1988). Chomsky and Herman begin by proclaiming that


4.5: Logarithmic Properties
In this section, you will:
- Use the Product Rule for logarithms.
- Use the Quotient Rule for logarithms.
- Use the Power Rule for logarithms.
- Expand logarithmic expressions.
- Condense logarithmic expressions.
- Use the Change-of-Base Formula for logarithms.
Figure \( \PageIndex{ 1 } \): The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)
In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from \( 0 \) to \( 14 \). Substances with a pH less than \( 7 \) are considered acidic, and substances with a pH greater than \( 7 \) are said to be alkaline. Our bodies, for instance, must maintain a pH close to \( 7.35 \) in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:
- Battery acid: 0.8
- Stomach acid: 2.7
- Orange juice: 3.3
- Pure water: 7 (at \(25^{\circ}\) C)
- Human blood: 7.35
- Fresh coconut: 7.8
- Sodium hydroxide (lye): 14
To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where \(H^+\), is the concentration of hydrogen ion in the solution\[ \text{pH} = -\log([H^+]) = \log\left( \dfrac{1}{[H^+]} \right). \nonumber \]The equivalence of \(-\log([H^+])\) and \( \log \left( \frac{1}{[H^+]}\right)\), is one of the Laws of Logarithms we will examine in this section.
Using the Product Rule for Logarithms
Recall that the logarithmic and exponential functions "undo" each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.\[ \log_b 1 = 0 \quad \text{and} \quad \log_b b = 1. \nonumber \]For example, \( \log_5 1 = 0\), since \( 5^0 = 1 \), and \( \log_5 5 = 1\), since \( 5^1 = 5 \).
Next, we have the inverse properties.\[ \log_b ( b^x ) = x \quad \text{and} \quad b^{\log_b x} = x \nonumber \]The second inverse property holds only if \( x > 0 \). To illustrate these properties, we imagine evaluating \( \log(100)\). We can rewrite the logarithm as \( \log_{10}( 10^2)\), and then apply the inverse property \( \log_b ( b^x ) = x\), to get \( \log_{10}(10^2) = 2\).
To evaluate \( e^{\ln( 7 )}\), we can rewrite the logarithm as \(e^{\log_e 7}\), and then apply the inverse property \(b^{\log_b x} =x\) to get \(e^{\log_e 7} = 7\).
Finally, we have the one-to-one property.\[ \log_b M = \log_b N \text{ if and only if } M = N. \nonumber \]We can use the one-to-one property to solve the equation \( \log_3 (3x) = \log_3 (2x+5)\), for \(x\). Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for \(x\):\[ \begin{array}{rrclcl} & 3x & = & 2x + 5 & \quad & \left( \text{setting the arguments equal to each other} \right) \\[6pt] \implies & x & = & 5 & \quad & \left( \text{subtracting }2x\text{ from both sides} \right) \\[6pt] \end{array} \nonumber \]But what about the equation \( \log_3 (3x) + \log_3 (2x+5) = 2\)? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.
Recall that we use the Product Rule of Exponents to combine the product of powers by adding exponents: \(x^a x^b = x^{a + b} \). We have a similar property for logarithms, called the Product Rule for Logarithms , which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the Product Rule below.
Given any real number \(x\), and positive real numbers \(M\), \(N\), and \(b\), where \(b \neq 1\), we will show\[ \log_b(MN) = \log_b (M) + \log_b (N). \nonumber \]Let \(m = \log_b M\), and \(n = \log_b N\). In exponential form, these equations are \(b^m = M\) and \(b^n = N\). It follows that\[ \begin{array}{rclcl} \log_b(MN) & = & \log_b(b^m b^n) & \quad & \left( \text{substituting} \right) \\[6pt] & = & \log_b(b^{m + n}) & \quad & \left( \text{Laws of Exponents: Product Rule} \right) \\[6pt] & = & m + n & \quad & \left( \text{Laws of Logarithms: Inverse Property} \right) \\[6pt] & = & \log_b(M) + \log_b(N) & \quad & \left( \text{substituting} \right) \\[6pt] \end{array} \nonumber \]Note that repeated applications of the Product Rule for Logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider \( \log_b (wxyz)\). Using the Product Rule for Logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:\[ \log_b (wxyz) = \log_b w + \log_b x + \log_b y + \log_b z. \nonumber \]
The Product Rule for Logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.\[ \log_b (MN) = \log_b (M) + \log_b (N) \nonumber \]for \( b > 0 \).
Expand \( \log_3 ( 30x ( 3x+4 ) )\).
- Solution
-
We begin by recognizing the argument as a product of three factors.\[ \log_3 ( 30x( 3x+4 ) )= \log_3 ( 30 \cdot x \cdot ( 3x+4 ) ). \nonumber \]Next we write the equivalent equation by summing the logarithms of each factor.\[ \log_3 ( 30 \cdot x \cdot ( 3x+4 ) )= \log_3 ( 30 ) + \log_3 ( x )+ \log_3 ( 3x+4 ). \nonumber \]Finally, we need to be aware of the base and the possibiliy that the \( 30 \) can be factored into a power of the base and some other constant. Specifically, \( 30 = 3 \cdot 10 \). Therefore,\[ \begin{array}{rclcl} \log_3 ( 30x( 3x+4 ) ) & = & \log_3 ( 3 \cdot 10 ) + \log_3 ( x )+ \log_3 ( 3x+4 ) & & \\[6pt] & = & \log_3 ( 3 ) + \log_3 ( 10 ) + \log_3 ( x )+ \log_3 ( 3x+4 ) & \quad & \left( \text{Laws of Logorithms: Product Rule} \right) \\[6pt] & = & 1 + \log_3 ( 10 ) + \log_3 ( x )+ \log_3 ( 3x+4 ) & \quad & \left( \text{Laws of Logorithms: }\log_b b = 1 \right) \\[6pt] \end{array} \nonumber \]
Expand \( \log_b (8k)\).
Using the Quotient Rule for Logarithms
For quotients, we have a similar rule for logarithms. Recall that we use the Quotient Rule of Exponents to combine the quotient of exponents by subtracting: \( \frac{x^a}{x^b} = x^{a−b} \). The Quotient Rule for Logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the Product Rule, we can use the inverse property to derive the Quotient Rule.
Given any real number \(x\), and positive real numbers \(M\), \(N\), and \(b\), where \(b \neq 1\), we will show\[ \log_b \left( \dfrac{M}{N} \right) = \log_b ( M )− \log_b ( N ). \nonumber \]Let \(m= \log_b M\), and \(n= \log_b N\). In exponential form, these equations are \( b^m = M \), and \(b^n = N\). It follows that\[ \begin{array}{rclcl} \log_b \left( \dfrac{M}{N} \right) & = & \log_b \left( \dfrac{b^m}{b^n} \right) & \quad & \left( \text{substituting} \right) \\[6pt] & = & \log_b \left( b^{m - n} \right) & \quad & \left( \text{Laws of Exponents: Quotient Rule} \right) \\[6pt] & = & m - n & \quad & \left( \text{Laws of Exponents: Inverse Property} \right) \\[6pt] & = & \log_b (M) - \log_b (N) & \quad & \left( \text{substituting} \right) \\[6pt] \end{array} \nonumber \]For example, to expand \( \log \left( \frac{2x^2 + 6x}{3x + 9} \right)\), we must first express the quotient in lowest terms. Factoring and canceling we get,\[ \begin{array}{rclcl} \log \left( \dfrac{2x^2 + 6x}{3x + 9} \right) & = & \log \left( \dfrac{2x(x + 3)}{3(x + 3)} \right) & \quad & \left( \text{factoring out the GCF from numerator and denominator} \right) \\[6pt] & = & \log \left( \dfrac{2x \cancel{(x + 3)}}{3\cancel{(x + 3)}} \right) & \quad & \left( \text{canceling like factors} \right) \\[6pt] & = & \log \left( \dfrac{2x}{3} \right) & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]Next we apply the Quotient Rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the Product Rule.\[ \log\left( \dfrac{2x}{3} \right) = \log(2x) - \log(3) = \log(2) + \log(x) - \log(3). \nonumber \]
The Quotient Rule for Logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.\[ \log_b \left( \dfrac{M}{N} \right)= \log_b M − \log_b N. \nonumber \]
Expand\[ \log_2 \left( \dfrac{15x(x - 1)}{(3x + 4)(2 - x)} \right). \nonumber \]
- Solution
-
First we note that the quotient is factored and in lowest terms, so we apply the Quotient Rule.\[ \log_2 \left( \dfrac{15x(x - 1)}{(3x + 4)(2 - x)} \right) = \log_2 ( 15x(x−1) ) − \log_2 ( (3x+4)(2−x) ). \nonumber \]Notice that the resulting terms are logarithms of products. To expand completely, we apply the Product Rule, noting that \( 15 \) is not divisible by a power of the base (\( 2 \)).\[ \begin{array}{rclcl} \log_2 ( 15x(x−1) ) − \log_2 ( (3x+4)(2−x) ) & = & \log_2 (15) + \log_2 (x) + \log_2(x−1) − \left[ \log_2(3x+4) + \log_2(2−x) \right] & \quad & \left( \text{Product Rule} \right) \\[6pt] & = & \log_2 (15) + \log_2 (x) + \log_2(x−1) − \log_2(3x+4) - \log_2(2−x) & \quad & \left( \text{distributing} \right) \\[6pt] \end{array} \nonumber \]
There are exceptions to consider in Example \( \PageIndex{ 2 } \) and later examples. First, because denominators must never be zero, the expression in Example \( \PageIndex{ 2 } \) is not defined for \(x= −\frac{4}{3}\) and \(x=2\). Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that \(x > 0\), \(x >1 \), \(x > −\frac{4}{3} \), and \(x < 2\). Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.
Expand \( \log_3 \left( \frac{7x^2 + 21x}{7x(x - 1)(x - 2)} \right)\).
Using the Power Rule for Logarithms
We’ve explored the Product Rule and the Quotient Rule, but how can we take the logarithm of a power, such as \( x^2 \)? One method is as follows:\[ \begin{array}{rclcl} \log_b(x^2) & = & \log_b(x \cdot x) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] & = & \log_b x + \log_b x & \quad & \left( \text{Laws of Logarithms: Product Rule} \right) \\[6pt] & = & 2 \log_b x & \quad & \left( \text{combining like terms} \right) \\[6pt] \end{array} \nonumber \]Notice that we used the Product Rule for Logarithms to find a solution for the example above. By doing so, we have derived the Power Rule for Logarithms , which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,\[100 = 10^2 \quad \sqrt{3} = 3^{1/2} \quad \dfrac{1}{e} = e^{−1}. \nonumber \]
The Power Rule for Logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.\[ \log_b ( M^n ) = n \log_b M \nonumber \]
Expand \( \log_2 x^5 \).
- Solution
-
The argument is already written as a power, so we identify the exponent, \( 5 \), and the base, \(x\), and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.\[ \log_2 ( x^5 ) =5 \log_2 x \nonumber \]
Expand \(\ln (x^2) \).
Expand \( \log_3 ( 25 )\), using the Power Rule for Logs.
- Solution
-
Expressing the argument as a power, we get \( \log_3 ( 25 )= \log_3 ( 5^2 )\).
Next we identify the exponent, \( 2 \), and the base, \( 5 \), and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.\[ \log_3 ( 5^2 ) = 2 \log_3 ( 5 ) \nonumber \]
Expand \(\ln\left( \frac{1}{x^2} \right)\).
Rewrite \(4\ln(x)\), using the Power Rule for logs to a single logarithm with a leading coefficient of \( 1 \).
- Solution
-
Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression \(4\ln(x)\), we identify the factor, \( 4 \), as the exponent and the argument, \(x\), as the base, and rewrite the product as a logarithm of a power: \(4 \ln(x)= \ln( x^4 )\).
Rewrite \(2 \log_3 4\), using the Power Rule for logs to a single logarithm with a leading coefficient of 1.
Expanding Logarithmic Expressions
Taken together, the Product Rule, Quotient Rule, and Power Rule are often called "Laws of Logarithms." Sometimes we apply more than one rule in order to simplify an expression. For example:\[ \begin{array}{rcl} \log_b\left( \dfrac{6x}{y} \right) & = & \log_b(6x) - \log_b y \\[6pt] & = & \log_b 6 + \log_b x - \log_b y \\[6pt] \end{array} \nonumber \]We can use the Power Rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the Quotient Rule for Logarithms using the fact that a reciprocal is a negative power:\[ \begin{array}{rcl} \log_b\left( \dfrac{A}{C} \right) & = & \log_b(A C^{-1}) \\[6pt] & = & \log_b (A) + \log_b (C^{-1}) \\[6pt] & = & \log_b A + (-1) \log_b C \\[6pt] & = & \log_b A - \log_b C \\[6pt] \end{array} \nonumber \]We can also apply the Product Rule to express a sum or difference of logarithms as the logarithm of a product.
With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots — never with addition or subtraction inside the argument of the logarithm.
Rewrite \(\ln\left( \frac{x^4 y}{7} \right)\), as a sum or difference of logs.
- Solution
-
First, because we have a quotient of two expressions, we can use the Quotient Rule:\[\ln\left( \dfrac{x^4 y}{7} \right)= \ln( x^4 y ) − \ln(7). \nonumber \]Then seeing the product in the first term, we use the Product Rule:\[ \ln( x^4 y ) − \ln(7) = \ln( x^4 ) + \ln(y) − \ln(7). \nonumber \]Finally, we use the Power Rule on the first term:\[\ln( x^4 ) + \ln(y) − \ln(7) = 4 \ln(x) + \ln(y) − \ln(7). \nonumber \]
Expand \( \log\left( \frac{x^2 y^3}{z^4} \right)\).
Expand \( \log( \sqrt{ x } )\).
- Solution
-
\[ \log( \sqrt{ x } ) = \log x^{1/2} = \dfrac{1}{2} \log x \nonumber \]
Expand \(\ln( \sqrt[3]{x^2} )\).
Can we expand \(\ln( x^2 + y^2 )\)?
No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.
Expand \( \log_6 \left( \frac{64x^3 (4x + 1)}{2x - 1} \right)\).
- Solution
-
We can expand by applying the Product and Quotient Rules.\[ \begin{array}{rcl} \log_6 \left( \frac{64x^3 (4x + 1)}{2x - 1} \right) & = & \log_6 64 + \log_6 x^3 + \log_6 (4x + 1) - \log_6 (2x - 1) \\[6pt] & = & \log_6 64 + 3\log_6 x + \log_6 (4x + 1) - \log_6 (2x - 1) \\[6pt] \end{array} \nonumber \]Note that we could have rewritten \( \log_6 64 \) as \( \log_6 2^6 = 6 \log_6 2 \), however, this is not any better than \( \log_6 64 \). When it comes to logarithms of constants, we are only concerned with the possibility that a power of the base is a factor of the constant.
Expand \( \ln\left( \frac{\sqrt{(x-1)(2x + 1)^2}}{x^2 - 9} \right)\).
Condensing Logarithmic Expressions
We can use the Laws of Logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined . We will learn later how to change the base of any logarithm before condensing. Because students sometimes make mistakes on the order with which they apply the Laws of Logarithms, we provide the following "How To" as a beginner's reference guide.
Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.
- Apply the Power Rule first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
- Next apply the Product Rule. Rewrite sums of logarithms as the logarithm of a product.
- Apply the Quotient Rule last. Rewrite differences of logarithms as the logarithm of a quotient.
Write \( \log_3 ( 5 )+ \log_3 ( 8 )− \log_3 ( 2 )\) as a single logarithm.
- Solution
-
Using the product and Quotient Rules\[ \log_3 ( 5 )+ \log_3 ( 8 )= \log_3 ( 5 \cdot 8 )= \log_3 ( 40 ). \nonumber \]This reduces our original expression to\[ \log_3 (40)− \log_3 (2). \nonumber \]Then, using the Quotient Rule\[ \log_3 ( 40 )− \log_3 ( 2 )= \log_3 \left( \dfrac{40}{2} \right)= \log_3 ( 20 ). \nonumber \]
Condense \( \log 3 - \log 4 + \log 5 - \log 6\).
Condense \( \log_2 ( x^2 )+ \frac{1}{2} \log_2 ( x−1 )−3 \log_2 ( ( x+3 )^2 )\).
- Solution
-
We apply the Power Rule first:\[ \log_2 ( x^2 )+ \frac{1}{2} \log_2 ( x−1 )−3 \log_2 ( ( x+3 )^2 )= \log_2 ( x^2 )+ \log_2 ( \sqrt{ x−1 } )− \log_2 ( ( x+3 )^6 ). \nonumber \]Next we apply the Product Rule to the sum:\[ \log_2 ( x^2 )+ \log_2 ( \sqrt{ x−1 } )− \log_2 ( ( x+3 )^6 )= \log_2 ( x^2 \sqrt{ x−1 } )− \log_2 ( ( x+3 )^6 ). \nonumber \]Finally, we apply the Quotient Rule to the difference:\[ \log_2 ( x^2 \sqrt{ x−1 } )− \log_2 ( ( x+3 )^6 )= \log_2 \left( \dfrac{x^2 \sqrt{ x−1 }}{( x+3 )^6} \right). \nonumber \]
Rewrite \( \log( 5 ) + 0.5 \log( x ) - \log( 7x−1 ) + 3\log( x−1 )\) as a single logarithm.
Rewrite \(2 \log x − 4 \log(x+5) + \frac{1}{x} \log( 3x+5 )\) as a single logarithm.
- Solution
-
We apply the Power Rule first:\[2 \log x – 4 \log(x+5) + \dfrac{1}{x} \log(3x+5) = \log( x^2 ) - \log ( x+5 )^4 + log\left( (3x+5)^{x^{−1}} \right). \nonumber \]Next, we rearrange and apply the Product Rule to the sum:\[ \begin{array}{rcl} \log( x^2 ) - \log ( x+5)^4 + \log\left( (3x+5)^{x^{−1}} \right) & = & \log( x^2 ) + \log\left( (3x+5)^{x^{−1}} \right) - \log ( x+5)^4 \\[6pt] & = & \log\left(x^2 (3x+5)^{x^{−1}} \right) - \log(x+5)^4 \\[6pt] \end{array} \nonumber \]Finally, we apply the Quotient Rule to the difference:\[ = \log \left( x^2 ( 3 x + 5 )^{x^{−1}} \right) − \log ( x + 5 )^4 = \log \dfrac{x^2 ( 3 x + 5 )^{x^{−1}}}{( x + 5 )^4}. \nonumber \]
Condense \(4( 3 \log( x ) + \log( x+5 ) - \log( 2x+3 ) )\).
Recall that, in chemistry, \(\text{pH} = -\log[ H^+ ]\). If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?
- Solution
-
Suppose \(C\), is the original concentration of hydrogen ions, and \(P\), is the original pH of the liquid. Then \(P = –\log(C)\). If the concentration is doubled, the new concentration is \(2C\). Then the pH of the new liquid is\[\text{pH}=-\log( 2C ). \nonumber \]Using the Product Rule of logs\[\text{pH} = -\log( 2C ) = −( \log(2) + \log(C) ) = -\log(2) - \log(C). \nonumber \]Since \(P = –\log(C)\), the new pH is\[\text{pH} = P - \log(2) \approx P − 0.301. \nonumber \]When the concentration of hydrogen ions is doubled, the pH decreases by about \( 0.301 \).
How does the pH change when the concentration of positive hydrogen ions is decreased by half?
Using the Change-of-Base Formula for Logarithms
Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than \( 10 \) or \(e\), we use the Change-of-Base Formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.
To derive the Change-of-Base Formula, we use the one-to-one property and Power Rule for logarithms.
Given any positive real numbers \(M\), \(b\), and \(n\), where \(n \neq 1\), and \(b \neq 1\), we show\[ \log_b M = \dfrac{\log_n M}{\log_n b}. \nonumber \]Let \(y= \log_b M\).By exponentiating both sides with base\(b\) we arrive at an exponential form, namely \(b^y =M\). It follows that\[ \begin{array}{rrclcl} & \log_n ( b^y ) & = & \log_n M & \quad & \left( \text{applying the One-to-One Property} \right) \\[6pt] \implies & y \log_n b & = & \log_n M & \quad & \left( \text{Laws of Logarithms: Power Rule} \right) \\[6pt] \implies & y & = & \dfrac{\log_n M}{\log_n b} & \quad & \left( \text{dividing both sides by }\log_n b \right) \\[6pt] \implies & \log_b M & = & \dfrac{\log_n M}{\log_n b} & \quad & \left( \text{substituting} \right) \\[6pt] \end{array} \nonumber \]For example, to evaluate \( \log_5 36\) using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.\[ \log_5 36 = \dfrac{\log( 36 )}{\log( 5 )} \approx 2.2266. \nonumber \]
For any positive real numbers \(M\), \(b\), and \(n\), where \(n \neq 1\), and \(b \neq 1\),\[ \log_b M = \dfrac{\log_n M}{\log_n b}. \nonumber \]It follows that the Change-of-Base Formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.\[ \log_b M = \dfrac{\ln M}{\ln b} \quad \text{and} \quad \log_b M = \dfrac{\log M}{\log b}. \nonumber \]
Change \( \log_5 3\) to a quotient of natural logarithms.
- Solution
-
Because we will be expressing \( \log_5 3\) as a quotient of natural logarithms, the new base, \(n=e\).
We rewrite the log as a quotient using the Change-of-Base Formula. The numerator of the quotient will be the natural log with argument \( 3 \). The denominator of the quotient will be the natural log with argument \( 5 \).\[ \log_5 3 = \dfrac{\ln 3}{\ln 5}. \nonumber \]
Change \( \log_{0.5} 8\) to a quotient of natural logarithms.
Can we change common logarithms to natural logarithms?
Yes. Remember that \( \log 9\), means \( \log_{10} 9\). So, \( \log 9 = \frac{\ln 9}{\ln 10} \).
Evaluate \( \log_2 (10)\), using the Change-of-Base Formula with a calculator.
- Solution
-
According to the Change-of-Base Formula, we can rewrite the log base \( 2 \) as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base \(e\).\[ \log_2 10 = \dfrac{\ln 10}{\ln 2} \approx 3.3219 . \nonumber \]
Evaluate \( \log_5 (100)\), using the Change-of-Base Formula.
Access these online resources for additional instruction and practice with laws of logarithms.


1.14.8: Calculus
Consider a variable u defined by the independent variables \(x\) and \(y\).
\[\text { We write } u=u[x, y] \nonumber \]
Equation (b) is the general exact differential of equation (a).
\[\mathrm{du}=\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}\right)_{\mathrm{y}} \, \mathrm{dx}+\left(\frac{\partial \mathrm{u}}{\partial \mathrm{y}}\right)_{\mathrm{x}} \, \mathrm{dy} \nonumber \]
In other words the change in u is related to the differential dependence of \(\mathrm{u}\) on \(x\) at constant \(y\) and the differential dependence of \(\mathrm{u}\) on \(y\) at constant \(x\). For the case where u does not change,
\[\left(\frac{\partial u}{\partial x}\right)_{y}=-\left(\frac{\partial u}{\partial y}\right)_{x} \,\left(\frac{\partial y}{\partial x}\right)_{u}=0 \text { and }\left(\frac{\partial y}{\partial x}\right)_{u}=-\left(\frac{\partial u}{\partial x}\right)_{y} \,\left(\frac{\partial y}{\partial u}\right)_{x} \nonumber \]
A variable \(z\) is defined by the independent variables \(x\) and \(y\).
\[z=z[x, y] \nonumber \]
Equation (e) is the general differential of equation (d). d z=\left(\frac{\partial z}{\partial x}\right)_{y} \, d x+\left(\frac{\partial z}{\partial y}\right)_{x} \, d y \nonumber \]
We direct attention to the dependence of \(z\) on \(x\) along a pathway for which \(\mathrm{u}\) is constant.
\[\text { Then }\left(\frac{\partial z}{\partial x}\right)_{u}=\left(\frac{\partial z}{\partial x}\right)_{y}+\left(\frac{\partial z}{\partial y}\right)_{x} \,\left(\frac{\partial y}{\partial x}\right)_{u} \nonumber \]
The latter equation contains the differential dependence of \(y\) on \(x\) at constant \(\mathrm{u}\). The latter dependence can be reformulated using equation (c). Therefore
\[\left(\frac{\partial z}{\partial x}\right)_{u}=\left(\frac{\partial z}{\partial x}\right)_{y}-\left(\frac{\partial u}{\partial x}\right)_{y} \,\left(\frac{\partial y}{\partial u}\right)_{x} \,\left(\frac{\partial z}{\partial y}\right)_{x} \nonumber \]
The key point to emerge from this exercise centres is the way in which the condition on the partial differential \((\partial z / \partial x)\) can be changed from ‘at constant \(y\)’ to ‘at constant \(\mathrm{u}\)’.
Another important operation concerns a variable \(\mathrm{q}\).
\[\text { Thus, }\left(\frac{\partial x}{\partial y}\right)_{z}=\left(\frac{\partial x}{\partial q}\right)_{z} \,\left(\frac{\partial q}{\partial y}\right)_{z} \nonumber \]
For composite functions such as \(z=z[\mathrm{u}, \mathrm{v}]\), where \(z=z[x, y]\), and \(\mathrm{u}=\mathrm{u}[\mathrm{x}, \mathrm{y}]\), further important equations are found [1].
\[\text { Thus }\left(\frac{\partial z}{\partial u}\right)_{v}=\left(\frac{\partial z}{\partial x}\right)_{y} \,\left(\frac{\partial x}{\partial u}\right)_{v}+\left(\frac{\partial z}{\partial y}\right)_{x} \,\left(\frac{\partial y}{\partial u}\right)_{v} \nonumber \]
Equation (i) is an example of the well-known chain rule, a similar equation holding for \((\partial z / \partial v)_{u}\). This rule allows the total change of independent variables from \(z=z[\mathrm{u}, \mathrm{v}]\) to \(z=z[x, y]\).
\[\text { Also }\left(\frac{\partial z}{\partial x}\right)_{y}=\left(\frac{\partial z}{\partial x}\right)_{y, v}+\left(\frac{\partial z}{\partial v}\right)_{y, x} \,\left(\frac{\partial v}{\partial x}\right)_{y} \nonumber \]
The latter equation is useful for introducing an extra constraint on a given differential.
Footnote
[1] H. B. Callen, Thermodynamics and an Introduction to Thermostatics, Wiley, New York, 2dn. Edn.,1985, Appendix A.