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Section 14.3 Using the Change-of-Base Formula

To evaluate a logarithm with any base other than \(10\) or \(e\text{,}\) we can use the Change-of-Base Formula. Understanding why this formula works is beyond the scope of this course, but may be covered, along with many other properties of logarithms, in your next college algebra course.

The Change-of-Base Formula introduces a new base \(a\text{.}\) This can be any base \(a\) we want where \(a\gt 0, a\neq 1\text{.}\) Because our calculators have keys for logarithms base \(10\) and base \(e\text{,}\) we will typically choose to write the Change-of-Base Formula with the new base as either \(10\) or \(e\text{.}\)

Subsection 14.3.1 The Change-of-Base Formula

Definition 14.3.1. Change-of-Base Formula.

For any logarithmic bases \(a\text{,}\) \(b\text{,}\) \(M\gt 0\) and \(\neq 1\text{,}\)

\begin{gather*} \mathrm{log}_{b}M=\frac{\mathrm{log}_{a}M}{\mathrm{log}_{a}b}\\ \\ \text{new base } a \end{gather*}
\begin{gather*} \mathrm{log}_{b}M=\frac{\highlight{\mathrm{log}}M}{\highlight{\mathrm{log}}b}\\ \\ \text{new base } 10 \end{gather*}
\begin{gather*} \mathrm{log}_{b}M=\frac{\highlight{\mathrm{ln}}M}{\highlight{\mathrm{ln}}b}\\ \\ \text{new base } e \end{gather*}

As you can see, you can choose to use either the "log" button or the "ln" button on your calculator to evaluate (find the value of) a logarithm with a base other than \(10\) or \(e\text{.}\) The important thing to remember is that you divide the log of the input by the log of the base and that you must use the same log button for both.

When we use a calculator to find the logarithm value, using the change-of-base formula, we will usually need to round our answers to a certain number of decimal places. This gives us an approximate value for the logarithm and so we use the approximately equal symbol (\(\approx\)).

Example 14.3.2.

Rounding to three decimal places, approximate \(\mathrm{log}_{4}35\text{.}\)

Explanation

We want to evaluate \(\mathrm{log}_{4}35\text{,}\) which is not base \(10\) or base \(e\text{.}\) So we can't enter it directly into our calculator at this point, but we can use the Change-of-Base Formula to rewrite it into either base \(10\) or base \(e\text{.}\) Then, we will be able to enter it into the calculator.

According to the Change-of-Base Formula,

\begin{equation*} \mathrm{log}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}\text{ or }\mathrm{log}_{b}M=\frac{\mathrm{ln}M}{\mathrm{ln}b} \end{equation*}

To use this formula, we need to identify that \(b=4\) and \(M=35\text{.}\)

Now let's choose log base \(10\) to rewrite the log.

\begin{equation*} \mathrm{log}_{4}35=\frac{\mathrm{log}35}{\mathrm{log}4}\text{.} \end{equation*}

Enter the expression \(\frac{\mathrm{log}35}{\mathrm{log}4}\) in the calculator using the log button for base \(10\text{.}\)

Then, rounding to three decimal places, we have

\begin{equation*} \mathrm{log}_{4}35\approx 2.565\text{.} \end{equation*}
Example 14.3.3.

Rounding to three decimal places, approximate \(\mathrm{log}_{6}0.52\text{.}\)

Explanation

We want to evaluate \(\mathrm{log}_{6}0.52\text{,}\) which is not base \(10\) or base \(e\text{.}\) So we can't enter it directly into our calculator at this point, but we can use the Change-of-Base Formula to rewrite it into either base \(10\) or base \(e\text{.}\) Then, we will be able to enter it into the calculator.

According to the Change-of-Base Formula,

\begin{equation*} \mathrm{log}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}\text{ or }\mathrm{log}_{b}M=\frac{\mathrm{ln}M}{\mathrm{ln}b} \end{equation*}

To use this formula, we need to identify that \(b=6\) and \(M=0.52\text{.}\)

This time let's choose log base \(e\) to rewrite the log.

\begin{equation*} \mathrm{log}_{6}0.52=\frac{\mathrm{ln}0.52}{\mathrm{ln}6}\text{.} \end{equation*}

Enter the expression \(\frac{\mathrm{ln}0.52}{\mathrm{ln}6}\) in the calculator using the ln button for base \(e\text{.}\)

Rounding to three decimal places, we have

\begin{equation*} \mathrm{log}_{6}0.52\approx -0.365 \end{equation*}

In the last example, what does the answer tell you? Remember that the log function is used to find the exponent to which the base must be raised to get the input as a result. So, this tells us that if we raise the base of \(6\) to the exponent of \(-0.365\text{,}\) then we should get a result that is approximately \(0.52\text{.}\) Note that it will only be approximate since we rounded our exponent. Let's check this result in the next example.

Example 14.3.4.

Verify that \(6^{-0.365}\approx 0.52\text{.}\)

Explanation

Entering \(6^{-0.365}\) into the calculator, we obtain \(0.5199658\text{...}\text{,}\) which of course is very close to \(0.52\text{.}\) This verifies our answer in the previous example.

Checkpoint 14.3.5.

Exercises 14.3.2 Exercises

In the following exercises, use the Change-of-Base Formula, rounding to three decimal places, to approximate each logarithm.

1.

\(\mathrm{log}_{8}2{,}500 \approx\)

2.

\(\mathrm{log}_{3}42\approx\)

3.

\(\mathrm{log}_{5}78 \approx\)

4.

\(\mathrm{log}_{12}87 \approx\)

5.

\(\mathrm{log}_{15}93 \approx\)

6.

\(\mathrm{log}_{2}17 \approx\)

7.

\(\mathrm{log}_{6}0.21 \approx\)

8.

\(\mathrm{log}_{7}120 \approx\)

This section is adapted from "Using the Change-of-Base Formula to Evaluate Logarithms" 1  by Wendy Lightheart, OpenStax CNX, which is licensed under CC BY 4.0 2 / A derivative from the original work "Use the Properties of Logarithms" 3  by OpenStax, OpenStax CNX.

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https://cnx.org/contents/tGunCymg@3/Use-the-Properties-of-Logarithms