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# Properties of Logarithms

### Learning Objectives

• The product rule for logarithms
• Define properties of logarithms, and use them to solve equations
• Define the product rule for logarithms, and use it to solve equations
• Quotient and Power Rules for Logarithms
• Define the quotient and power rules for logarithms
• Use the quotient and power rules for logarithms to simplify logarithmic expressions
•  Expand and Condense Logarithms
• Combine product, power and quotient rules to simplify logarithmic expressions
• Expand logarithmic expressions that have negative or fractional exponents
• Condense logarithmic expressions
• Change of Base
• Use properties of logarithms to define the change of base formula
• Change the base of logarithmic expressions into base 10, or base e

When you learned how to solve linear equations, you were likely introduced to the properties of real numbers. These properties help us know what the rules are for isolating and combining numbers and variables. For example, it is advantageous to know that multiplication and division "undo" each other when you want to solve an equation for a variable that is multiplied by a number. You may have also been introduced to properties and rules for writing and using exponents. In this section, we will show you properties of logarithms and how they can help you better understand what a logarithm means and eventually how to solve equations that contain them.

Recall that we can express the relationship between logarithmic form and its corresponding exponential form as follows:

${\mathrm{log}}_{b}\left(x\right)=y\Leftrightarrow {b}^{y}=x,\text{}b>0,b\ne 1$

Note that the base b is always positive and 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, we will introduce some basic properties of logarithms followed by examples with integer arguments to help you get familiar with the relationship between exponents and logarithms.

### Zero and Identity Exponent Rule for Logarithms and Exponentials

${\mathrm{log}}_{b}1=0$, b>0
${\mathrm{log}}_{b}b=1$, b>0

### Example

Use the the fact that exponentials and logarithms are inverses to prove the zero and identity exponent rule for the following:

1.${\mathrm{log}}_{5}1=0$

2. ${\mathrm{log}}_{5}5=1$

1.${\mathrm{log}}_{5}1=0$  since ${5}^{0}=1$

2.${\mathrm{log}}_{5}5=1$ since ${5}^{1}=5$

Exponential and logarithmic functions are inverses of each other and we can take advantage of this to evaluate and solve expressions and equations involving logarithms and exponentials. The inverse property of logarithms and exponentials gives us an explicit way to rewrite an exponential as a logarithm or a logarithm as an exponential.

### Inverse Property of Logarithms and Exponentials

$\begin{array}{c}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x, x>0, b>0, b\ne1\hfill \end{array}$

### Example

Evaluate: 1.$\mathrm{log}\left(100\right)$ 2.${e}^{\mathrm{ln}\left(7\right)}$

Answer: 1.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 ${\mathrm{log}}_{10}\left({10}^{2}\right)=2$. 2.Rewrite the logarithm as ${e}^{{\mathrm{log}}_{e}7}$, and then apply the inverse property ${b}^{{\mathrm{log}}_{b}x}=x$ to get ${e}^{{\mathrm{log}}_{e}7}=7$

Another property that can help us simplify logarithms is the one-to-one property. Essentially, this property states that if two logarithms that have the same base are equal to each other, then their arguments - the stuff inside - is also equal to each other.

### The One-To-One Property of Logarithms

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

### Example

Solve the equation ${\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)$ for x.

Answer: In order for this equation to be true we must find a value for x such that $3x=2x+5$

$\begin{array}{c}3x=2x+5\hfill & \text{Set the arguments equal}\text{.}\hfill \\ x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}$
$\begin{array}{c}{\mathrm{log}}_{3}\left(3\cdot5\right)={\mathrm{log}}_{3}\left(2\cdot5+5\right)\\{\mathrm{log}}_{3}\left(15\right)={\mathrm{log}}_{3}\left(15\right)\end{array}$
This is a true statement, so we must have found the correct value for x.

What if we had started with 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 terms on the left side of the equation. To recap - the properties of logarithms and exponentials that can help us understand, simplify and solve these types of functions more easily include:
• Zero and Identity Exponent Rule: ${\mathrm{log}}_{b}1=0$, b>0, and ${\mathrm{log}}_{b}b=1$, b>0
• Inverse Property: $\begin{array}{c}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}$
• One-To-One Property: ${\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N$

## 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.

### 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.

${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\text{ for }b>0$
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
$\begin{array}{c}{\mathrm{log}}_{b}\left(MN\right)\hfill & ={\mathrm{log}}_{b}\left({b}^{m}{b}^{n}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m+n}\right)\hfill & \text{Apply the product rule for exponents}.\hfill \\ \hfill & =m+n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}$

Repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. Consider the following example:

### Example

Using the product rule for logarithms, rewrite this logarithm of a product as the sum of logarithms of its factors. [latex-display]{\mathrm{log}}_{b}\left(wxyz\right)[/latex-display]

In our next example, we will first factor the argument of a logarithm before expanding it with the product rule.

### Example

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

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

${\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)={\mathrm{log}}_{3}\left(2\cdot 3\cdot 5\cdot x\cdot \left(3x+4\right)\right)$

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

${\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)={\mathrm{log}}_{3}\left(2\right)+{\mathrm{log}}_{3}\left(3\right)+{\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)$

### Analysis of the Solution

It is tempting to use the distributive property when you see an expression like $\left(30x\left(3x+4\right)\right)$, but in this case, it is better to leave the argument of this logarithm as a product since you can then use the product rule for logarithms to simplify the expression.

## 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.

### 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.

${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$
We can show ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{=}{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$
Given positive real numbers M, N, and b, where $b>0$ 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

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

### Example

Expand the following expression using the quotient rule for logarithms. [latex-display]\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)[/latex-display]

Factoring and canceling we get,

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

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.

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

In the previous example, it was helpful to first factor the numerator and denominator and divide common terms.  This gave us a simpler expression to use to write an equivalent expression. It is important to remember to subtract the logarithm of the denominator from the logarithm of the numerator. Always check to see if you need to expand further with the product rule.

### Example

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

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

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

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.

$\begin{array}{c}{\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right) \\\text{ }= \left[{\mathrm{log}}_{2}\left(3\right)+{\mathrm{log}}_{2}\left(5\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)\right]-\left[{\mathrm{log}}_{2}\left(3x+4\right)+{\mathrm{log}}_{2}\left(2-x\right)\right]\hfill \\ \text{ }={\mathrm{log}}_{2}\left(3\right)+{\mathrm{log}}_{2}\left(5\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)-{\mathrm{log}}_{2}\left(3x+4\right)-{\mathrm{log}}_{2}\left(2-x\right)\hfill \end{array}$

## Analysis of the Solution

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 = 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that > 0, > 1, $x>-\frac{4}{3}$, and < 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:

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

Notice that we used the product rule for logarithms to simplify 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,

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

### 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.

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

### Example

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

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.

${\mathrm{log}}_{2}\left({x}^{5}\right)=5{\mathrm{log}}_{2}x$

The power rule for logarithms is possible because we can use the product rule and combine like terms. In the next example you will see that we can also rewrite an expression as a power in order to use the power rule.

### Example

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

Expressing the argument as a power, we get ${\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right)$.

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.

${\mathrm{log}}_{3}\left({5}^{2}\right)=2{\mathrm{log}}_{3}\left(5\right)$

Now let's use the power rule in reverse.

### Example

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

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 $4\mathrm{ln}\left(x\right)$, 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:

$4\mathrm{ln}\left(x\right)=\mathrm{ln}\left({x}^{4}\right)$.

## Expand and Condense Logarithms

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:

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

We can also 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 (fraction) has a negative power:

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

We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.

Remember that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm. Consider the following example:

$\begin{array}{c}\mathrm{log}\left(10+100\right)\overset{?}{=}\end{array}\mathrm{log}\left(10\right)+\mathrm{log}\left(100\right)\\\mathrm{log}\left(110\right)\overset{?}{=}1+2\\2.04\ne3$

Be careful to only apply the product rule when a logarithm has an argument that is a product or when you have a sum of logarithms.
In our first example we will show that a logarithmic expression can be expanded by combining several of the rules of logarithms.

### Example

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

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

$\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)=\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)$

Then seeing the product in the first term, we use the product rule:

$\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)=\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)$

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

$\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)=4\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)$

We can also use the rules for logarithms to simplify the logarithm of a radical expression.

### Example

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

Answer: $\begin{array}{c}\mathrm{log}\left(\sqrt{x}\right)\hfill & =\mathrm{log}{x}^{\left(\frac{1}{2}\right)}\hfill \\ \hfill & =\frac{1}{2}\mathrm{log}x\hfill \end{array}$

Can we expand $\mathrm{ln}\left({x}^{2}+{y}^{2}\right)$? Use the textbox below to develop an argument one way or the other before you look at the solution.[practice-area rows="1"][/practice-area]

Answer: No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Rewrite the expression as an equation and express it as an exponential to give yourself some proof. [latex-display]m=\mathrm{ln}\left({x}^{2}+{y}^{2}\right)[/latex-display] If you rewrite this as an exponential you get: [latex-display]e^m={x}^{2}+{y}^{2}[/latex-display] From here, there's not much more you can do to make this expression more simple.

Let's do one more example with an expression that contains several different mathematical operations.

### Example

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

We can expand by applying the Product and Quotient Rules.

$\begin{array}{c}{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)\hfill & ={\mathrm{log}}_{6}64+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Quotient and Product Rules}.\hfill \\ \hfill & ={\mathrm{log}}_{6}{2}^{6}+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & {\text{Simplify by writing 64 as 2}}^{6}.\hfill \\ \hfill & =6{\mathrm{log}}_{6}2+3{\mathrm{log}}_{6}x+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Power Rule}.\hfill \end{array}$

## Condense 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.

### Example

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

Using the product and quotient rules

${\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)={\mathrm{log}}_{3}\left(5\cdot 8\right)={\mathrm{log}}_{3}\left(40\right)$

This reduces our original expression to

${\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)$

Then, using the quotient rule

${\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)={\mathrm{log}}_{3}\left(\frac{40}{2}\right)={\mathrm{log}}_{3}\left(20\right)$

In our next example, we show how to simplify a more complex logarithm by condensing it.

### Example

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

We apply the power rule first:

${\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)={\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)$

Next we apply the product rule to the sum:

${\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)$

Finally, we apply the quotient rule to the difference:

${\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\frac{{x}^{2}\sqrt{x - 1}}{{\left(x+3\right)}^{6}}$

## Change of Base

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 $n\ne 1$ and $b\ne 1$, we show

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

Let $y={\mathrm{log}}_{b}M$. By taking the log base $n$ of both sides of the equation, we arrive at an exponential form, namely ${b}^{y}=M$. It follows that

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

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.

$\begin{array}{c}{\mathrm{log}}_{5}36\hfill & =\frac{\mathrm{log}\left(36\right)}{\mathrm{log}\left(5\right)}\hfill & \text{Apply the change of base formula using base 10}\text{.}\hfill \\ \hfill & \approx 2.2266\text{ }\hfill & \text{Use a calculator to evaluate to 4 decimal places}\text{.}\hfill \end{array}$
Let's practice changing the base of a logarithmic expression from 5 to base e.

### Example

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

Because we will be expressing ${\mathrm{log}}_{5}3$ as a quotient of natural logarithms, the new base, = 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.

$\begin{array}{c}{\mathrm{log}}_{b}M\hfill & =\frac{\mathrm{ln}M}{\mathrm{ln}b}\hfill \\ {\mathrm{log}}_{5}3\hfill & =\frac{\mathrm{ln}3}{\mathrm{ln}5}\hfill \end{array}$

We can generalize the change of base formula in the following way:

### 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

${\mathrm{log}}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}$
As we stated earlier, the main reason for changing the base of a logarithm is to be able to evaluate it with a calculator. In the following example we will use the change of base formula on a logarithmic expression, then evaluate the result with a calculator.

### Example

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

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.

$\begin{array}{c}{\mathrm{log}}_{2}10=\frac{\mathrm{ln}10}{\mathrm{ln}2}\hfill & \text{Apply the change of base formula using base }e.\hfill \\ \approx 3.3219\hfill & \text{Use a calculator to evaluate to 4 decimal places}.\hfill \end{array}$

Can we change common logarithms to natural logarithms? Write your ideas in the textbox below before looking at the solution. [practice-area rows="1"][/practice-area]

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

## Summary

Logarithms have properties that can help us simplify and solve expressions and equations that contain logarithms. Exponentials and logarithms are inverses of each other, therefore we can define the product rule for logarithms. We can use this as follows to simplify or solve expressions with logarithms.  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.

You can use the quotient rule of logarithms to write an equivalent difference of logarithms in the following way:

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.

To use the power rule of logarithms to write an equivalent product of a factor and a logarithm consider the following:

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

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. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.
For practical purposes found in many different sciences or finance applications, you may want to evaluate a logarithm with a calculator. The change of base formula will allow you to change the base of any logarithm to either 10 or  so you can evaluate it with a calculator. Here we have summarized the steps for using the change of base formula.

Given a logarithm with the form ${\mathrm{log}}_{b}M$

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.

## Why learn about exponential and logarithmic equations?

Joan enjoyed talking to her grandfather so much that the next Sunday, she made sure she would be there when he came to dinner. This time, though, she prepared a question for him.  She had been learning about logarithmic and exponential functions in her math class and wanted to see if she could stump her grandfather with a math question.
Kinemetrics Seismograph formerly used by the United States Department of the Interior.
One of the questions in Joan's homework on exponential and logarithmic functions had been about how to calculate the Richter scale measure of the magnitude of an earthquake. The following formula was given in her book:

$R=\mathrm{log}\left(\frac{A}{A_{0}}\right)$

The question provided her with the quantity $\frac{A}{A_{0}}$ and asked her to solve for R.  ${A_{0}}$ is a baseline measure of ground movement as detected by a seismometer, seen in the image above. When there is an earthquake, wave amplitudes recorded by seismometers are expressed relative to this baseline.  For example, Joan's book asked the following question:

An earthquake is measured with a wave amplitude 392 times as great as $A_{0}$. What is the magnitude of this earthquake using the Richter scale, to the nearest tenth?

Joan hoped to give her grandfather the Richter scale magnitude for the Alaska quake, 8.5, and see if he could find how much greater the wave amplitude of that quake was than the baseline, ${A_{0}}$.

In this module you will learn how to solve problems such as the one Joan is planning to try to stump her grandfather with. We will come back to Joan and her grandfather at the end of this module to see if she was able to ask him a question he didn't know how to answer.

In this module, you will learn about the properties of exponential and logarithmic functions in the same way that you learned about the properties of exponents.  You will use the properties of logarithms and exponentials to solve equations that involve them.

### The learning outcomes for this module include:

• Define and use the properties of logarithms to expand, condense, and change the base of a logarithmic expression
• Use the properties of logarithms and exponentials to solve equations