How to Change Base Formula for Logarithms?

Changing the base formula, as the name suggests, is used to change the base of the logarithm. You may have noticed that a scientific calculator only has “\(log\)” and “\(ln\)” buttons.

We know that “\(log\) ” stands for a logarithm of base \(10\) and “\(ln\)” stands for a logarithm of base \(e\). But there is no option to calculate the logarithm of a number with bases other than \(10\) and \(e\). Changing the base formula solves this problem.

Related Topics

• Properties of Logarithms
• How to Solve Natural Logarithms Problems

A step-by-step guide to the change of base formula for logarithms

The change of base formula is used to write a logarithm of a number with a definite base as the ratio of two logarithms each with the same base that is different from the base of the original logarithm. This is the property of logarithms.

Change of base formula

The change of base formula is:

\(\color{blue}{log _b\left(a\right)=\frac{\left[log _c\left(a\right)\right]}{\left[log _c\left(b\right)\right]}}\)

In this formula:

• The logarithm argument in the numerator is the same as the original logarithm argument.
• The logarithm argument at the denominator is the same as the base of the original logarithm.
• The base of both numerator and denominator logarithms must be the same, and this base can be any positive number other than \(1\).

Note: Another form of this formula is, \(log _b\left(a\right). log _c\left(b\right)=log _c\left(a\right)\), which is also widely used in solving problems.

The Change of Base Formula for Logarithms – Example 1:

Evaluate the value of \(log _{64}\left(8\right)\) using the change of base formula.

Solution:

Apply the base formula change by changing the base to \(10\):

\(log _{64}\left(8\right)=\frac{log\:8}{log\:64}\)

\(=\frac{log\:8}{log\:8^2}\)

\(=\frac{log\:8}{2\:log\:8}\)

\(=\frac{1}{2}\)

Exercises for the Change of Base Formula for Logarithms

Evaluate each logarithm by using the change of base formula.

1. \(\color{blue}{log _{\frac{1}{4}}\left(32\sqrt{2}\right)}\)
2. \(\color{blue}{log _{16}\left(32\right)}\)
3. \(\color{blue}{log _{\frac{1}{27}}\left(243\right)}\)
4. \(\color{blue}{log _{0.1}\left(1000\right)}\)

1. \(\color{blue}{-\frac{11}{4}}\)
2. \(\color{blue}{\frac{5}{4}}\)
3. \(\color{blue}{-\frac{5}{3}}\)
4. \(\color{blue}{-3}\)

Recommended EffortlessMath Books

For a workbook that covers logarithms inside a full Algebra II program, Algebra II for Beginners walks through log rules, change of base, and exponential equations with worked examples. For test-prep coverage, The Ultimate Algebra Bundle covers logs alongside polynomials, rational functions, and conic sections.

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Frequently Asked Questions

Why do we need the change-of-base formula?

Because calculators only have buttons for two specific log bases: \(\log\) (base \(10\)) and \(\ln\) (base \(e\)). Anything else — \(\log_5\), \(\log_2\), \(\log_{17}\) — has to be rewritten before the calculator can handle it.

What is the change-of-base formula?

\(\log_b(a) = \dfrac{\log_c(a)}{\log_c(b)}\), where \(c\) is any positive base other than \(1\). You’re converting from base \(b\) to base \(c\).

Which new base should I pick?

Whichever your calculator has. Base \(10\) (\(\log\)) is the most common choice. Base \(e\) (\(\ln\)) is preferred in calculus. Both give the same final answer.

Can the new base be anything?

Almost. It can be any positive real number except \(1\). \(\log_1\) is undefined because \(1\) raised to any power is still \(1\). In practice, only \(10\) and \(e\) get used.

How do I evaluate \(\log_2(8)\) with the formula?

\(\log_2(8) = \dfrac{\log 8}{\log 2}\). On a calculator: \(\dfrac{0.903}{0.301} = 3\). Or recognize that \(2^3 = 8\), so \(\log_2(8) = 3\) directly.

Does the formula work with \(\ln\) instead of \(\log\)?

Yes. \(\log_b(a) = \dfrac{\ln a}{\ln b}\). The same conversion holds for any base on the bottom and top.

What’s the alternate form of the change-of-base formula?

\(\log_b(a) \cdot \log_c(b) = \log_c(a)\). It’s the same identity written differently and shows up in proofs of other log properties.

Can I use the change-of-base formula to solve exponential equations?

Yes. To solve \(5^x = 20\), take \(\log\) of both sides: \(x \log 5 = \log 20\), so \(x = \dfrac{\log 20}{\log 5} = \log_5(20) \approx 1.861\). The change-of-base formula is hiding inside.

What’s a common mistake with this formula?

Putting the argument and base in the wrong spots. The argument \(a\) goes on top, the original base \(b\) goes on bottom. Swapping them flips the answer to its reciprocal.

Where does this show up on tests?

Algebra II finals, Precalculus, ACT Math, SAT Math (occasionally), AP Calculus, College Algebra placement tests, and ALEKS. Any test with logs has at least one problem that needs change of base.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

• Properties of logarithms
• How to solve natural logarithm problems
• How to solve exponential equations
• Rules of exponents
• How to use the quadratic formula