## Properties of Logarithms

### Learning Outcomes

• Rewrite a logarithmic expression using the power rule, product rule, or quotient rule.
• Expand logarithmic expressions using a combination of logarithm rules.
• Condense logarithmic expressions using logarithm rules.
• Expand a logarithm using a combination of logarithm rules.
• Condense a logarithmic expression into one logarithm.
• Rewrite logarithms with a different base using the change of base formula.

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° 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 $a$ is the concentration of hydrogen ion in the solution.

$\begin{array}{lll}\text{pH} & = -\mathrm{log}\left(\left[{H}^{+}\right]\right)\hfill \\ \text{} & =\mathrm{log}\left(\frac{1}{\left[{H}^{+}\right]}\right)\hfill \end{array}$

$-\mathrm{log}\left(\left[{H}^{+}\right]\right)$ is equal to $\mathrm{log}\left(\frac{1}{\left[{H}^{+}\right]}\right)$ due to one of the logarithm properties we will examine in this section.

## Properties of 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.

$\begin{array}{l}{\mathrm{log}}_{b}1=0\\{\mathrm{log}}_{b}b=1\end{array}$

For example, ${\mathrm{log}}_{5}1=0$ since ${5}^{0}=1$ and ${\mathrm{log}}_{5}5=1$ since ${5}^{1}=5$.

Next, we have the inverse property.

$\begin{array}{l}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}$

For example, to evaluate $\mathrm{log}\left(100\right)$, we can rewrite the logarithm as ${\mathrm{log}}_{10}\left({10}^{2}\right)$ and then apply the inverse property ${\mathrm{log}}_{b}\left({b}^{x}\right)=x$ to get ${\mathrm{log}}_{10}\left({10}^{2}\right)=2$.

To evaluate ${e}^{\mathrm{ln}\left(7\right)}$, we can rewrite the logarithm as ${e}^{{\mathrm{log}}_{e}7}$ and then apply the inverse property ${b}^{{\mathrm{log}}_{b}x}=x$ to get ${e}^{{\mathrm{log}}_{e}7}=7$.

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

${\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N$

We can use the one-to-one property to solve the equation ${\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)$ 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:

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

But what about the equation ${\mathrm{log}}_{3}\left(3x\right)+{\mathrm{log}}_{3}\left(2x+5\right)=2$? 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 logarithms on the left side of the equation.

### Using the Product Rule for Logarithms

Recall that we use the product rule of exponents to combine the product of exponents by adding: ${x}^{a}{x}^{b}={x}^{a+b}$. 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 $b\ne 1$, we will show

${\mathrm{log}}_{b}\left(MN\right)\text{= }{\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$.

Let $m={\mathrm{log}}_{b}M$ and $n={\mathrm{log}}_{b}N$. In exponential form, these equations are ${b}^{m}=M$ and ${b}^{n}=N$. It follows that

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

### A General Note: The Product Rule for Logarithms

The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.

${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\text{ for }b>0$

### Example: Using the Product Rule for Logarithms

Expand ${\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)$.

### Try It

Expand ${\mathrm{log}}_{b}\left(8k\right)$.

### 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: ${x}^{\frac{a}{b}}={x}^{a-b}$. 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 and positive real numbers M, N, and b, where $b\ne 1$, we will show

${\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{= }{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$.

Let $m={\mathrm{log}}_{b}M$ and $n={\mathrm{log}}_{b}N$. In exponential form, these equations are ${b}^{m}=M$ and ${b}^{n}=N$. It follows that

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

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

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

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.

$\begin{array}{lll}\mathrm{log}\left(\frac{2x}{3}\right) & =\mathrm{log}\left(2x\right)-\mathrm{log}\left(3\right)\hfill \\ \text{} & =\mathrm{log}\left(2\right)+\mathrm{log}\left(x\right)-\mathrm{log}\left(3\right)\hfill \end{array}$

### A General Note: 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.

${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$

### 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: Using the Quotient Rule for Logarithms

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

### Try It

Expand ${\mathrm{log}}_{3}\left(\frac{7{x}^{2}+21x}{7x\left(x - 1\right)\left(x - 2\right)}\right)$.

### Using the Power Rule for Logarithms

We have 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:

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

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,

$\begin{array}{lll}100={10}^{2}, \hfill & \sqrt{3}={3}^{\frac{1}{2}}, \hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}$

### A General Note: 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.

${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$

### 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: Expanding a Logarithm with Powers

Rewrite ${\mathrm{log}}_{2}{x}^{5}$.

### Try It

Rewrite $\mathrm{ln}{x}^{2}$.

### Example: Rewriting an Expression as a Power before Using the Power Rule

Rewrite ${\mathrm{log}}_{3}\left(25\right)$ using the power rule for logs.

### Try It

Rewrite $\mathrm{ln}\left(\frac{1}{{x}^{2}}\right)$.

## Expanding Logarithms

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

$\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{6x}{y}\right)\hfill & ={\mathrm{log}}_{b}\left(6x\right)-{\mathrm{log}}_{b}y\hfill \\ \hfill & ={\mathrm{log}}_{b}6+{\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y\hfill \end{array}$

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:

$\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{A}{C}\right)\hfill & ={\mathrm{log}}_{b}\left(A{C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(A\right)+{\mathrm{log}}_{b}\left({C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}A+\left(-1\right){\mathrm{log}}_{b}C\hfill \\ \hfill & ={\mathrm{log}}_{b}A-{\mathrm{log}}_{b}C\hfill \end{array}$

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 and then just 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: Using a combination of the rules for logarithms to expand a logarithm

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

### Try It

Expand $\mathrm{log}\left(\frac{{x}^{2}{y}^{3}}{{z}^{4}}\right)$.

In the next example we will recall that we can write roots as exponents, and use this quality to simplify logarithmic expressions.

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

Expand $\mathrm{log}\left(\sqrt{x}\right)$.

### Try It

Expand $\mathrm{ln}\left(\sqrt[3]{{x}^{2}}\right)$.

### Q & A

Can we expand $\mathrm{ln}\left({x}^{2}+{y}^{2}\right)$?

No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.

Now we will provide some examples that will require careful attention.

### Example: Expanding Complex Logarithmic Expressions

Expand ${\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)$.

### Try It

Expand $\mathrm{ln}\left(\frac{\sqrt{\left(x - 1\right){\left(2x+1\right)}^{2}}}{\left({x}^{2}-9\right)}\right)$.

## Condensing Logarithms

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.

### How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm

1. Apply the power property first. Identify terms that are products of factors and a logarithm and rewrite each as the logarithm of a power.
2. From left to right, apply the product and quotient properties. Rewrite sums of logarithms as the logarithm of a product and differences of logarithms as the logarithm of a quotient.

### Example: Using the Power Rule in Reverse

Use the power rule for logs to rewrite $4\mathrm{ln}\left(x\right)$ as a single logarithm with a leading coefficient of 1.

### Try It

Use the power rule for logs to rewrite $2{\mathrm{log}}_{3}4$ as a single logarithm with a leading coefficient of 1.

In our next few examples we will use a combination of logarithm rules to condense logarithms.

### Example: Using the Product and Quotient Rules to Combine Logarithms

Write ${\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)-{\mathrm{log}}_{3}\left(2\right)$ as a single logarithm.

### Try It

Condense $\mathrm{log}3-\mathrm{log}4+\mathrm{log}5-\mathrm{log}6$.

### Example: Condensing Complex Logarithmic Expressions

Condense ${\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)$.

### Example: Rewriting as a Single Logarithm

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

### Try It

Rewrite $\mathrm{log}\left(5\right)+0.5\mathrm{log}\left(x\right)-\mathrm{log}\left(7x - 1\right)+3\mathrm{log}\left(x - 1\right)$ as a single logarithm.

Condense $4\left(3\mathrm{log}\left(x\right)+\mathrm{log}\left(x+5\right)-\mathrm{log}\left(2x+3\right)\right)$.

### Applications of Properties of Logarithms

In chemistry, pH is a measure of how acidic or basic a liquid is. It is essentially a measure of the concentration of hydrogen ions in a solution. The scale for measuring pH is standardized across the world, the scientific community having agreed upon its values and methods for acquiring them.

Measurements of pH can help scientists, farmers, doctors, and engineers solve problems and identify sources of problems.

pH is defined as the decimal logarithm of the reciprocal of the hydrogen ion activity, $a_{H}+$, in a solution.
$\text{pH} =-\log _{10}(a_{{\text{H}}^{+}})=\log _{10}\left({\frac {1}{a_{{\text{H}}^{+}}}}\right)$

For example, a solution with a hydrogen ion activity of $2.5×{10}^{-6}$ (at that level essentially the number of moles of hydrogen ions per liter of solution) has a pH of $\log_{10}\left(\frac{1}{2.5×{10}^{-6}}\right)=5.6$

In the next examples, we will solve some problems involving pH.

### Example: Applying Properties of Logs

Recall that, in chemistry, $\text{pH}=-\mathrm{log}\left[{H}^{+}\right]$. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

### Try It

How does the pH change when the concentration of positive hydrogen ions is decreased by half?

## Change-of-Base Formula for Logarithms

Most calculators can only evaluate 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 $n\ne 1$ and $b\ne 1$, we show

${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}$

Let $y={\mathrm{log}}_{b}M$. Converting to exponential form, we obtain ${b}^{y}=M$. It follows that:

$\begin{array}{l}{\mathrm{log}}_{n}\left({b}^{y}\right)\hfill & ={\mathrm{log}}_{n}M\hfill & \text{Apply the one-to-one property}.\hfill \\ y{\mathrm{log}}_{n}b\hfill & ={\mathrm{log}}_{n}M \hfill & \text{Apply the power rule for logarithms}.\hfill \\ y\hfill & =\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\hfill & \text{Isolate }y.\hfill \\ {\mathrm{log}}_{b}M\hfill & =\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\hfill & \text{Substitute for }y.\hfill \end{array}$

For example, to evaluate ${\mathrm{log}}_{5}36$ using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.

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

### A General Note: The Change-of-Base Formula

The change-of-base formula can be used to evaluate a logarithm with any base.

For any positive real numbers M, b, and n, where $n\ne 1$ and $b\ne 1$,

${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}$.

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.

${\mathrm{log}}_{b}M=\frac{\mathrm{ln}M}{\mathrm{ln}b}$

and

${\mathrm{log}}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}$

### How To: Given a logarithm Of the form ${\mathrm{log}}_{b}M$, use the change-of-base formula to rewrite it as a quotient of logs with any positive base $n$, where $n\ne 1$

1. Determine the new base n, remembering that the common log, $\mathrm{log}\left(x\right)$, has base 10 and the natural log, $\mathrm{ln}\left(x\right)$, has base e.
2. Rewrite the log as a quotient using the change-of-base formula:
• The numerator of the quotient will be a logarithm with base n and argument M.
• The denominator of the quotient will be a logarithm with base n and argument b.

### Example: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs

Change ${\mathrm{log}}_{5}3$ to a quotient of natural logarithms.

### Try It

Change ${\mathrm{log}}_{0.5}8$ to a quotient of natural logarithms.

### Q & A

Can we change common logarithms to natural logarithms?

Yes. Remember that $\mathrm{log}9$ means ${\text{log}}_{\text{10}}\text{9}$. So, $\mathrm{log}9=\frac{\mathrm{ln}9}{\mathrm{ln}10}$.

### Example: Using the Change-of-Base Formula with a Calculator

Evaluate ${\mathrm{log}}_{2}\left(10\right)$ using the change-of-base formula with a calculator.

### Try It

Evaluate ${\mathrm{log}}_{5}\left(100\right)$ using the change-of-base formula.

## Key Equations

 The Product Rule for Logarithms ${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$ The Quotient Rule for Logarithms ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$ The Power Rule for Logarithms ${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$ The Change-of-Base Formula ${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\text{ }n>0,n\ne 1,b\ne 1$

## Key Concepts

• We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms.
• We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms.
• 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.
• We can use the product rule, quotient rule, and power rule together to combine or expand a logarithm with a complex input.
• The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm.
• We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.
• The change-of-base formula is often used to rewrite a logarithm with a base other than 10 or $e$ as the quotient of natural or common logs. A calculator can then be used to evaluate it.

## Glossary

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