Logarithmic Properties

Learning Objectives

Simplify expressions using the properties for exponents. (IA 5.2.1)
Use the properties of logarithms. (IA 10.4.1)

Objective 1: Simplify expressions using the properties for exponents (IA 5.2.1)

The Product Property

Simplify expressions using the properties for exponents.

The figure shows how to multiply exponentials with the same base. In the example we start with x raised to the power of 2 times x raised to the power of 3. This means the we are multiplying 2 factors of x with 3 factors of x for a total of 5 factors of x so the simplified result is x raised to the power of 5.
Simplify		$x^{2} \cdot x^{3}$
What does this mean?
Now we see that		$x^{2} \cdot x^{3} = x^{5}$

To multiply powers with the same base we need to ________ exponents.

This leads us to the Product Property $a^{m} \cdot a^{n} = a^{m + n}$

The Quotient Property

Simplify $\frac{x^{5}}{x^{2}}$

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What does this mean?	$\frac{xxxxx}{xx}$	After simplifying we get $x^{3}$
Now we see that	$\frac{x^{5}}{x^{2}} = x^{3}$

To divide powers with the same base we need to __________ exponents.

This leads us to the Quotient Property $a^{m} \cdot a^{n} = a^{m - n}$

The Power Property

Simplify $(x^{2})^{4}$

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What does this mean?	$x^{2} \cdot x^{2} \cdot x^{2} \cdot x^{2}$	After adding exponents we get $x^{8}$ .
Now we see that	$(x^{2})^{4} = x^{8}$

To raise a power to a power we need to __________ exponents.

This leads us to the Power Property $(a^{m})^{n} = a^{m n}$ .

We will also use these other properties:

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Negative Exponents Property	$x^{- n} = \frac{1}{x^{n}}, x \neq 0$
Zero Exponent Property	$a^{0} = 1, i f a \neq 0$

Example 1

Simplify expressions using the properties for exponents.

ⓐ Simplify $3 ∙ 2^{x} ∙ 2^{3 x}$
ⓑ Simplify $\frac{b^{2} b^{6} b}{b^{4} b^{7}}$
ⓒ Simplify ${(a b^{2})}^{3} a^{5} b^{- 6}$

Solution

ⓐ

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Use the product property.	$3 ∙ 2^{x + 3 x}$
Simplify.	$3 ∙ 2^{4 x}$

ⓑ

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Use the product property and multiply exponents.	$\frac{b^{9}}{b^{11}}$
Use the quotient property and add exponents.	$b^{- 2} = \frac{1}{b^{2}}$

ⓒ

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Use the power property and multiply exponents.	$a^{3} b^{6} a^{5} b^{- 6}$
Use the product property and add exponents.	$a^{8} b^{0}$
Any base to the power of zero equals 1.	$a^{8} (1) = a^{8}$

Practice Makes Perfect

Simplify expressions using the properties for exponents.

$3 b^{5} ∙ 2 b^{12}$

$x ∙ x^{5} ∙ x^{7}$

$\frac{b^{15} c^{4}}{b^{4} c}$

$4 x^{0}$

$12 x^{-6}$

${(2 a^{3})}^{3} {(a^{4})}^{2}$

$\frac{a^{- 3} b^{5}}{2 a^{- 6}}$

Objective 2: Use the properties of logarithms (IA 10.4.1).

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Property	Base $a$	Base $e$
	${log}_{a} 1 = 0$	$ln 1 = 0$
	${log}_{a} a = 1$	$ln e = 1$
Inverse Properties	$\begin{matrix} a^{{log}_{a} x} = x \\ {log}_{a} a^{x} = x \end{matrix}$	$\begin{matrix} e^{ln x} = x \\ ln e^{x} = x \end{matrix}$
Product Property of Logarithms	${log}_{a} (M \cdot N) = {log}_{a} M + {log}_{a} N$	$ln (M \cdot N) = ln M + ln N$
Quotient Property of Logarithms	${log}_{a} \frac{M}{N} = {log}_{a} M - {log}_{a} N$	$ln \frac{M}{N} = ln M - ln N$
Power Property of Logarithms	${log}_{a} M^{p} = p {log}_{a} M$	$ln M^{p} = p ln M$

Example 2

Use the Properties of Logarithms to expand the logarithm ${log}_{4} (2 x^{3} y^{2})$ . Simplify, if possible.

Solution

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	${log}_{4} (2 x^{3} y^{2})$
Use the Product Property, ${log}_{a} M \cdot N = {log}_{a} M + {log}_{a} N$ .	${log}_{4} 2 + {log}_{4} x^{3} + {log}_{4} y^{2}$
Use the Power Property, ${log}_{a} M^{p} = p {log}_{a} M$ , on the last two terms.	${log}_{4} 2 + 3 {log}_{4} x + 2 {log}_{4} y$
Simplify.	$\frac{1}{2} + 3 {log}_{4} x + 2 {log}_{4} y$
	${log}_{4} (2 x^{3} y^{2}) = \frac{1}{2} + 3 {log}_{4} x + 2 {log}_{4} y$

Example 3

Use the Properties of Logarithms to expand the logarithm ${log}_{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}}$ . Simplify, if possible.

Solution

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	${log}_{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}}$
Rewrite the radical with a rational exponent.	${log}_{2} {(\frac{x^{3}}{3 y^{2} z})}^{\frac{1}{4}}$
Use the Power Property, ${log}_{a} M^{p} = p {log}_{a} M$ .	$\frac{1}{4} {log}_{2} (\frac{x^{3}}{3 y^{2} z})$
Use the Quotient Property, ${log}_{a} M \cdot N = {log}_{a} M - {log}_{a} N$ .	$\frac{1}{4} ({log}_{2} (x^{3}) - {log}_{2} (3 y^{2} z))$
Use the Product Property, ${log}_{a} M \cdot N = {log}_{a} M + {log}_{a} N$ , in the second term.	$\frac{1}{4} ({log}_{2} (x^{3}) - ({log}_{2} 3 + {log}_{2} y^{2} + {log}_{2} z))$
Use the Power Property, ${log}_{a} M^{p} = p {log}_{a} M$ , inside the parentheses.	$\frac{1}{4} (3 {log}_{2} x - ({log}_{2} 3 + 2 {log}_{2} y + {log}_{2} z))$
Simplify by distributing.	$\frac{1}{4} (3 {log}_{2} x - {log}_{2} 3 - 2 {log}_{2} y - {log}_{2} z)$
	${log}_{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}} = \frac{1}{4} (3 {log}_{2} x - {log}_{2} 3 - 2 {log}_{2} y - {log}_{2} z)$

Example 4

Use the Properties of Logarithms to condense the logarithm ${log}_{4} 3 + {log}_{4} x - {log}_{4} y$ . Simplify, if possible.

Solution

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The log expressions all have the same base, 4.	${log}_{4} 3 + {log}_{4} x - {log}_{4} y$
The first two terms are added, so we use the Product Property, ${log}_{a} M + {log}_{a} N = {log}_{a} M \cdot N$ .	${log}_{4} 3 x - {log}_{4} y$
Since the logs are subtracted, we use the Quotient Property, ${log}_{a} M - {log}_{a} N = {log}_{a} \frac{M}{N}$ .	${log}_{4} \frac{3 x}{y}$
	${log}_{4} 3 + {log}_{4} x - {log}_{4} y = {log}_{4} \frac{3 x}{y}$

Practice Makes Perfect

Use the properties of logarithms to expand: $\log_{3} (9 x^{5} y^{4})$

Use the properties of logarithms to expand: $\log_{5} \sqrt[3]{\frac{x^{5}}{25 y^{3} z}}$

Use the Properties of Logarithms to condense the logarithm: $\log_{b} 5 + \log_{b} c - \log_{b} b$

Use the Properties of Logarithms to condense the logarithm: $2 \log_{3} x + 3 \log_{3} (x + 1)$

Testing of the pH of hydrochloric acid. — Figure 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 basic. 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 basic, 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 basic, 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

\begin{array}{l} pH = - \log ([H^{+}]) \\ = \log (\frac{1}{[H^{+}]}) \end{array}

The equivalence of $- \log ([H^{+}])$ and $\log (\frac{1}{[H^{+}]})$ is one of the logarithm properties 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.

\begin{array}{l} \log_{b} 1 = 0 \\ \log_{b} b = 1 \end{array}

For example, $\log_{5} 1 = 0$ since $5^{0} = 1.$ And $\log_{5} 5 = 1$ since $5^{1} = 5.$

Next, we have the inverse property.

\begin{array}{l} \log_{b} (b^{x}) = x \\ b^{\log_{b} x} = x, x > 0 \end{array}

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 $e^{\log_{e} 7},$ 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 if and only if M = N

We can use the one-to-one property to solve the equation $\log_{3} (3 x) = \log_{3} (2 x + 5)$ 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}{l} 3 x = 2 x + 5 & Set the arguments equal . \\ x = 5 & Subtract 2 x . \end{array}

But what about the equation $\log_{3} (3 x) + \log_{3} (2 x + 5) = 2 ?$ 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} (M N) = \log_{b} (M) + \log_{b} (N) .

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}{l} \log_{b} (M N) & = \log_{b} (b^{m} b^{n}) & Substitute for M and N . \\ = \log_{b} (b^{m + n}) & Apply the product rule for exponents . \\ = m + n & Apply the inverse property of logs . \\ = \log_{b} (M) + \log_{b} (N) & Substitute for m and n . \end{array}

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 $\log_{b} (w x y z) .$ Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

\log_{b} (w x y z) = \log_{b} w + \log_{b} x + \log_{b} y + \log_{b} z

Example 5

Using the Product Rule for Logarithms

Expand $\log_{3} (30 x (3 x + 4)) .$

Solution

We begin by factoring the argument completely, expressing $30$ as a product of primes.

\log_{3} (30 x (3 x + 4)) = \log_{3} (2 \cdot 3 \cdot 5 \cdot x \cdot (3 x + 4))

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

\log_{3} (30 x (3 x + 4)) = \log_{3} (2) + \log_{3} (3) + \log_{3} (5) + \log_{3} (x) + \log_{3} (3 x + 4)

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} (\frac{M}{N}) = \log_{b} (M) - \log_{b} (N) .

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}{l} \log_{b} (\frac{M}{N}) & = \log_{b} (\frac{b^{m}}{b^{n}}) & Substitute for M and N . \\ = \log_{b} (b^{m - n}) & Apply the quotient rule for exponents . \\ = m - n & Apply the inverse property of logs . \\ = \log_{b} (M) - \log_{b} (N) & Substitute for m and n . \end{array}

For example, to expand $\log (\frac{2 x^{2} + 6 x}{3 x + 9}),$ we must first express the quotient in lowest terms. Factoring and canceling we get,

\begin{array}{l} \log (\frac{2 x^{2} + 6 x}{3 x + 9}) = \log (\frac{2 x (x + 3)}{3 (x + 3)}) & Factor the numerator and denominator . \\ = \log (\frac{2 x}{3}) & Cancel the common factors . \end{array}

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.

\begin{array}{l} \log (\frac{2 x}{3}) = \log (2 x) - \log (3) \\ = \log (2) + \log (x) - \log (3) \end{array}

Example 6

Using the Quotient Rule for Logarithms

Expand $\log_{2} (\frac{15 x (x - 1)}{(3 x + 4) (2 - x)}) .$

Solution

First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.

\log_{2} (\frac{15 x (x - 1)}{(3 x + 4) (2 - x)}) = \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}{l} \log_{2} (15 x (x - 1)) - \log_{2} ((3 x + 4) (2 - x)) = [\log_{2} (3) + \log_{2} (5) + \log_{2} (x) + \log_{2} (x - 1)] - [\log_{2} (3 x + 4) + \log_{2} (2 - x)] \\ = \log_{2} (3) + \log_{2} (5) + \log_{2} (x) + \log_{2} (x - 1) - \log_{2} (3 x + 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.

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}{l} \log_{b} (x^{2}) & = \log_{b} (x \cdot x) \\ = \log_{b} x + \log_{b} x \\ = 2 \log_{b} x \end{array}

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,

\begin{array}{l} 100 = 10^{2} & \sqrt{3} = 3^{\frac{1}{2}} & \frac{1}{e} = e^{- 1} \end{array}

Example 7

Expanding a Logarithm with Powers

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

Example 8

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}) = 2 \log_{3} (5)

Example 9

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

Expanding Logarithmic Expressions

Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example:

\begin{array}{l} \log_{b} (\frac{6 x}{y}) & = \log_{b} (6 x) - \log_{b} y \\ = \log_{b} 6 + \log_{b} x - \log_{b} y \end{array}

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}{l} \log_{b} (\frac{A}{C}) & = \log_{b} (A C^{- 1}) \\ = \log_{b} (A) + \log_{b} (C^{- 1}) \\ = \log_{b} A + (- 1) \log_{b} C \\ = \log_{b} A - \log_{b} C \end{array}

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.

Example 10

Expanding Logarithms Using Product, Quotient, and Power Rules

Rewrite $\ln (\frac{x^{4} y}{7})$ as a sum or difference of logs.

Solution

First, because we have a quotient of two expressions, we can use the quotient rule:

\ln (\frac{x^{4} y}{7}) = \ln (x^{4} y) - \ln (7)

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)

Finally, we use the power rule on the first term:

\ln (x^{4}) + \ln (y) - \ln (7) = 4 \ln (x) + \ln (y) - \ln (7)

Example 11

Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression

Expand $\log (\sqrt{x}) .$

Solution

\begin{array}{l} \log (\sqrt{x}) & = \log x^{(\frac{1}{2})} \\ = \frac{1}{2} \log x \end{array}

Example 12

Expanding Complex Logarithmic Expressions

Expand $\log_{6} (\frac{64 x^{3} (4 x + 1)}{(2 x - 1)}) .$

Solution

We can expand by applying the Product and Quotient Rules.

\begin{array}{l} \log_{6} (\frac{64 x^{3} (4 x + 1)}{(2 x - 1)}) & = \log_{6} 64 + \log_{6} x^{3} + \log_{6} (4 x + 1) - \log_{6} (2 x - 1) & Apply the Quotient Rule . \\ = \log_{6} 2^{6} + \log_{6} x^{3} + \log_{6} (4 x + 1) - \log_{6} (2 x - 1) & {Simplify by writing  64 as 2}^{6} . \\ = 6 \log_{6} 2 + 3 \log_{6} x + \log_{6} (4 x + 1) - \log_{6} (2 x - 1) & Apply the Power Rule . \end{array}

Condensing Logarithmic Expressions

We can use the rules 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.

Example 13

Using the Product and Quotient Rules to Combine Logarithms

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)

This reduces our original expression to

\log_{3} (40) - \log_{3} (2)

Then, using the quotient rule

\log_{3} (40) - \log_{3} (2) = \log_{3} (\frac{40}{2}) = \log_{3} (20)

Example 14

Condensing Complex Logarithmic Expressions

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})

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})

Finally, we apply the quotient rule to the difference:

\log_{2} (x^{2} \sqrt{x - 1}) - \log_{2} ({(x + 3)}^{6}) = \log_{2} \frac{x^{2} \sqrt{x - 1}}{{(x + 3)}^{6}}

Example 15

Rewriting as a Single Logarithm

Rewrite $2 \log x - 4 \log (x + 5) + \frac{1}{x} \log (3 x + 5)$ as a single logarithm.

Solution

We apply the power rule first:

2 \log x - 4 \log (x + 5) + \frac{1}{x} \log (3 x + 5) = \log (x^{2}) - \log {(x + 5)}^{4} + \log ({(3 x + 5)}^{x^{- 1}})

Next we rearrange and apply the product rule to the sum:

\log (x^{2}) - \log {(x + 5)}^{4} + \log ({(3 x + 5)}^{x^{- 1}})

= \log (x^{2}) + \log ({(3 x + 5)}^{x^{- 1}}) - \log {(x + 5)}^{4}

= \log (x^{2} {(3 x + 5)}^{x^{- 1}}) - \log {(x + 5)}^{4}

Finally, we apply the quotient rule to the difference:

= \log (x^{2} {(3 x + 5)}^{x^{−1}}) - {\log (x + 5)}^{4} = \log \frac{x^{2} {(3 x + 5)}^{x^{−1}}}{{(x + 5)}^{4}}

Example 16

Applying of the Laws of Logs

Recall that, in chemistry, $pH = - \log [H^{+}] .$ 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 $2 C .$ Then the pH of the new liquid is

pH = - \log (2 C)

Using the product rule of logs

pH = - \log (2 C) = - (\log (2) + \log (C)) = - \log (2) - \log (C)

Since $P = - \log (C),$ the new pH is

pH = P - \log (2) \approx P - 0.301

When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.

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 = \frac{\log_{n} M}{\log_{n} b}

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}{l} \log_{n} (b^{y}) & = \log_{n} M & Apply the one-to-one property . \\ y \log_{n} b & = \log_{n} M & Apply the power rule for logarithms. \\ y & = \frac{\log_{n} M}{\log_{n} b} & Isolate y . \\ \log_{b} M & = \frac{\log_{n} M}{\log_{n} b} & Substitute for y . \end{array}

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.

\begin{array}{l} \log_{5} 36 & = \frac{\log (36)}{\log (5)} & Apply the change of base formula using base 10 . \\ \approx 2.2266 & Use a calculator to evaluate to 4 decimal places . \end{array}

Example 17

Changing Logarithmic Expressions to Expressions Involving Only Natural Logs

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.

\begin{array}{l} \log_{b} M & = \frac{\ln M}{\ln b} \\ \log_{5} 3 & = \frac{\ln 3}{\ln 5} \end{array}

Example 18

Using the Change-of-Base Formula with a Calculator

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 $e .$

\begin{array}{l} \log_{2} 10 = \frac{\ln 10}{\ln 2} & Apply the change of base formula using base e . \\ \approx 3.3219 & Use a calculator to evaluate to 4 decimal places . \end{array}

Key Equations

...
The Product Rule for Logarithms	$\log_{b} (M N) = \log_{b} (M) + \log_{b} (N)$
The Quotient Rule for Logarithms	$\log_{b} (\frac{M}{N}) = \log_{b} M - \log_{b} N$
The Power Rule for Logarithms	$\log_{b} (M^{n}) = n \log_{b} M$
The Change-of-Base Formula	$\log_{b} M = \frac{\log_{n} M}{\log_{n} b} n > 0, n \neq 1, b \neq 1$

Key Concepts

We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See Example 5.
We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See Example 6.
We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See Example 7, Example 8, and Example 9.
We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See Example 10, Example 11, and Example 12.
The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See Example 13, Example 14, Example 15, and Example 16.
We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula. See Example 17.
The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and $e$ as the quotient of natural or common logs. That way a calculator can be used to evaluate. See Example 18.

Section Exercises

Verbal

Exercise 1

How does the power rule for logarithms help when solving logarithms with the form $\log_{b} (\sqrt[n]{x}) ?$

Solution

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, $\log_{b} (x^{\frac{1}{n}}) = \frac{1}{n} \log_{b} (x) .$

Exercise 2

What does the change-of-base formula do? Why is it useful when using a calculator?

Algebraic

For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

Exercise 3

$\log_{b} (7 x \cdot 2 y)$

Solution

$\log_{b} (2) + \log_{b} (7) + \log_{b} (x) + \log_{b} (y)$

Exercise 4

$\ln (3 a b \cdot 5 c)$

Exercise 5

$\log_{b} (\frac{13}{17})$

Solution

$\log_{b} (13) - \log_{b} (17)$

Exercise 6

$\log_{4} (\frac{\frac{x}{z}}{w})$

Exercise 7

$\ln (\frac{1}{4^{k}})$

Solution

$- k \ln (4)$

Exercise 8

$\log_{2} (y^{x})$

For the following exercises, condense to a single logarithm if possible.

Exercise 9

$\ln (7) + \ln (x) + \ln (y)$

Solution

$\ln (7 x y)$

Exercise 10

$\log_{3} (2) + \log_{3} (a) + \log_{3} (11) + \log_{3} (b)$

Exercise 11

$\log_{b} (28) - \log_{b} (7)$

Solution

$\log_{b} (4)$

Exercise 12

$\ln (a) - \ln (d) - \ln (c)$

Exercise 13

$- \log_{b} (\frac{1}{7})$

Solution

${log}_{b} (7)$

Exercise 14

$\frac{1}{3} \ln (8)$

For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

Exercise 15

$\log (\frac{x^{15} y^{13}}{z^{19}})$

Solution

$15 \log (x) + 13 \log (y) - 19 \log (z)$

Exercise 16

$\ln (\frac{a^{−2}}{b^{−4} c^{5}})$

Exercise 17

$\log (\sqrt{x^{3} y^{- 4}})$

Solution

$\frac{3}{2} \log (x) - 2 \log (y)$

Exercise 18

$\ln (y \sqrt{\frac{y}{1 - y}})$

Exercise 19

$\log (x^{2} y^{3} \sqrt[3]{x^{2} y^{5}})$

Solution

$\frac{8}{3} \log (x) + \frac{14}{3} \log (y)$

For the following exercises, condense each expression to a single logarithm using the properties of logarithms.

Exercise 20

$\log (2 x^{4}) + \log (3 x^{5})$

Exercise 21

$\ln (6 x^{9}) - \ln (3 x^{2})$

Solution

$\ln (2 x^{7})$

Exercise 22

$2 \log (x) + 3 \log (x + 1)$

Exercise 23

$\log (x) - \frac{1}{2} \log (y) + 3 \log (z)$

Solution

$\log (\frac{x z^{3}}{\sqrt{y}})$

Exercise 24

$4 \log_{7} (c) + \frac{\log_{7} (a)}{3} + \frac{\log_{7} (b)}{3}$

For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.

Exercise 25

$\log_{7} (15)$ to base $e$

Solution

$\log_{7} (15) = \frac{\ln (15)}{\ln (7)}$

Exercise 26

$\log_{14} (55.875)$ to base $10$

For the following exercises, suppose $\log_{5} (6) = a$ and $\log_{5} (11) = b .$ Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of $a$ and $b .$ Show the steps for solving.

Exercise 27

$\log_{11} (5)$

Solution

$\log_{11} (5) = \frac{\log_{5} (5)}{\log_{5} (11)} = \frac{1}{b}$

Exercise 28

$\log_{6} (55)$

Exercise 29

$\log_{11} (\frac{6}{11})$

Solution

$\log_{11} (\frac{6}{11}) = \frac{\log_{5} (\frac{6}{11})}{\log_{5} (11)} = \frac{\log_{5} (6) - \log_{5} (11)}{\log_{5} (11)} = \frac{a - b}{b} = \frac{a}{b} - 1$

Numeric

For the following exercises, use properties of logarithms to evaluate without using a calculator.

Exercise 30

$\log_{3} (\frac{1}{9}) - 3 \log_{3} (3)$

Exercise 31

$6 \log_{8} (2) + \frac{\log_{8} (64)}{3 \log_{8} (4)}$

Solution

$3$

Exercise 32

$2 \log_{9} (3) - 4 \log_{9} (3) + \log_{9} (\frac{1}{729})$

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

Exercise 33

$\log_{3} (22)$

Solution

$2.81359$

Exercise 34

$\log_{8} (65)$

Exercise 35

$\log_{6} (5.38)$

Solution

$0.93913$

Exercise 36

$\log_{4} (\frac{15}{2})$

Exercise 37

$\log_{\frac{1}{2}} (4.7)$

Solution

$- 2.23266$

Extensions

Exercise 38

Use the product rule for logarithms to find all $x$ values such that $\log_{12} (2 x + 6) + \log_{12} (x + 2) = 2.$ Show the steps for solving.

Exercise 39

Use the quotient rule for logarithms to find all $x$ values such that $\log_{6} (x + 2) - \log_{6} (x - 3) = 1.$ Show the steps for solving.

Solution

$x = 4;$ By the quotient rule: $\log_{6} (x + 2) - \log_{6} (x - 3) = \log_{6} (\frac{x + 2}{x - 3}) = 1.$

Rewriting as an exponential equation and solving for $x :$

$\begin{array}{l} 6^{1} & = \frac{x + 2}{x - 3} (after multiplying both sides by (x - 3)) \\ 6 (x - 3) & = x + 2 (carry out distributive multiplication on left) \\ 6 x - 18 & = x + 2 (add 18 to both sides, subtract x from both sides) \\ 5 x & = 20 \\ x & = 4 \end{array}$

Checking, we find that $\log_{6} (4 + 2) - \log_{6} (4 - 3) = \log_{6} (6) - \log_{6} (1)$ is defined, so $x = 4.$

Exercise 40

Can the power property of logarithms be derived from the power property of exponents using the equation $b^{x} = m ?$ If not, explain why. If so, show the derivation.

Exercise 41

Prove that $\log_{b} (n) = \frac{1}{\log_{n} (b)}$ for any positive integers $b > 1$ and $n > 1.$

Solution

Let $b$ and $n$ be positive integers greater than $1.$ Then, by the change-of-base formula, $\log_{b} (n) = \frac{\log_{n} (n)}{\log_{n} (b)} = \frac{1}{\log_{n} (b)} .$

Exercise 42

Does $\log_{81} (2401) = \log_{3} (7) ?$ Verify the claim algebraically.

change-of-base formula

a formula for converting a logarithm with any base to a quotient of logarithms with any other base.

power rule for logarithms

a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base

product rule for logarithms

a rule of logarithms that states that the log of a product is equal to a sum of logarithms

quotient rule for logarithms

a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms