Properties of Logarithms

Learning Objectives

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 1 The pH of hydrochloric acid is tested with litmus paper.

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° 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

$pH=−log([H^+])$

$=log \left(\frac{1}{[H^+]} \right)$

The equivalence of $−log([H^+])$ and $log \left(\frac{1}{[H^+]} \right)$ is one of the logarithm properties we will examine in this section.

Source: Rice University, https://openstax.org/books/college-algebra/pages/6-5-logarithmic-properties
This work is licensed under a Creative Commons Attribution 4.0 License.

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.

$\begin{aligned} log_b1=0 \\ log_bb=1 \end{aligned}$

Next, we have the inverse property.

$\begin{aligned} log_b(b^x) &=x \\ b^{log_bx} &=x,x > 0 \end{aligned}$

For example, to evaluate $log(100)$, 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 $e^{log_e7}$, and then apply the inverse property $b^{log_bx}=x$ to get $e^{log_e7}=7$.

Finally, we have the one-to-one property.

$log_bM = log_bN \quad \text { if and only if } M=N$

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 $x$:

$\begin{aligned} 3x=2x+5 & & &\text { Set the arguments equal }. \\ x=5 & & &\text { Subtract } 2x. \end{aligned}$

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: $x^ax^b=x^{a+b}$. 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

$log_b(MN)=log_b(M)+log_b(N)$.

Let $m=log_bM$ and $n=log_bN$. In exponential form, these equations are $b^m=M$ and $b^n=N$. It follows that

$\begin{aligned} \log _{b}(M N) &=\log _{b}\left(b^{m} b^{n}\right) & & \text { Substitute for } M \text { and } N\\ &=\log _{b}\left(b^{m+n}\right) & & \text { Apply the product rule for exponents. }\\ &=m+n & & \text { Apply the inverse property of logs. }\\ &=\log _{b}(M)+\log _{b}(N) & & \text { Substitute for } m \text { and } n \text {. } \end{aligned}$

$log_b(wxyz)=log_bw+log_bx+log_by+log_bz$

THE PRODUCT RULE FOR LOGARITHMS

$log_b(MN)=log_b(M)+log_b(N)$ for $b > 0$

HOW TO

Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.

1. Factor the argument completely, expressing each whole number factor as a product of primes.
2. Write the equivalent expression by summing the logarithms of each factor.

EXAMPLE 1

Using the Product Rule for Logarithms

Expand $log_3(30x(3x+4))$.

Solution

$log_3(30x(3x+4))=log_3(2⋅3⋅5⋅x⋅(3x+4))$

Next we write the equivalent equation by summing the logarithms of each factor.

$log_3(30x(3x+4))=log_3(2)+log_3(3)+log_3(5)+log_3(x)+log_3(3x+4)$

TRY IT #1

Expand $log_b(8k)$.

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

$log_b(\frac{M}{N})=log_b(M)−log_b(N)$.

Let $m=log_bM$ and $n=log_bN$. In exponential form, these equations are $b^m=M$ and $b^n=N$. It follows that

$\begin{aligned} \log _{b}\left(\frac{M}{N}\right) &=\log _{b}\left(\frac{b^{m}}{b^{n}}\right) & & \text { Substitute for } M \text { and } N . \\ &=\log _{b}\left(b^{m-n}\right) & & \text { Apply the quotient rule for exponents. } \\ &=m-n & & \text { Apply the inverse property of logs. } \\ &=\log _{b}(M)-\log _{b}(N) & & \text { Substitute for } m \text { and } n . \end{aligned}$

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,

$\begin{aligned} log \left(\frac{2x^2+6x}{3x+9}\right) &= log \left(\frac{2x(x+3)}{3(x+3)} \right) & & \text { Factor the numerator and denominator. } \\ &=log \left( \frac{2x}{3} \right) & & \text { Cancel the common factors. } \end{aligned}$

$\begin{aligned} log (\frac{2x}{3}) &= log(2x)−log(3) \\ &=log(2)+log(x)−log(3) \end{aligned}$

THE QUOTIENT RULE FOR LOGARITHMS

The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.

$log_b(\frac{M}{N})=log_bM−log_bN$

HOW TO

Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.

1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

EXAMPLE 2

Using the Quotient Rule for Logarithms

Expand $log_2 \left(\frac{15x(x−1)}{(3x+4)(2−x)}\right)$.

Solution

$\log _{2}\left(\frac{15 x(x-1)}{(3 x+4)(2-x)}\right)=\log _{2}(15 x(x-1))-\log _{2}((3 x+4)(2-x))$

Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.

$\begin{array}{ll} log_2(15x(x−1))−log_2((3x+4)(2−x)) &=[log_2(3)+log_2(5)+log_2(x)+log_2(x−1)]−[log_2(3x+4)+log_2(2−x)] \\ &=log_2(3)+log_2(5)+log_2(x)+log_2(x−1)−log_2(3x+4)−log_2(2−x) \end{array}$

Analysis

There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression 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.

TRY IT #2

Expand $log_3 \left(\frac{7x^2+21x}{7x(x−1)(x−2)}\right)$.

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:

$\begin{aligned} \log _{b}\left(x^{2}\right) &=\log _{b}(x \cdot x) \\ &=\log _{b} x+\log _{b} x \\ &=2 \log _{b} x \end{aligned}$

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,

$100=10^2 \quad \sqrt{3}=3 \frac{1}{2} \quad \frac{1}{e}=e^{−1}$

THE POWER RULE FOR LOGARITHMS

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.

$log_b(M^n)=nlog_bM$

HOW TO

Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.

1. Express the argument as a power, if needed.
2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.

EXAMPLE 3

Expanding a Logarithm with Powers

Expand $log_2x^5$.

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.

$log_2(x^5)=5log_2x$

TRY IT #3

Expand $\ln x^2$.

EXAMPLE 4

Rewriting an Expression as a Power before Using the Power Rule

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.

$log_3(5^2)=2log_3(5)$

TRY IT #4

Expand $\ln (\frac{1}{x^2})$.

EXAMPLE 5

Using the Power Rule in Reverse

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: $4 \ln (x) = \ln(x^4)$.

TRY IT #5

Rewrite $2log_34$ using the power rule for logs to a single logarithm with a leading coefficient of 1.