3.4.4: Change of Base

Change of Base

While it is possible to change bases by always going back to exponential form, it is more efficient to find out how to change the base of logarithms in general. Since there are only base e and base 10 logarithms on most calculators, how would you evaluate an expression like log₃12?

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Changing the Base of Logarithms

The change of base property states:

\(\ \log _{b} x=\frac{\log _{a} x}{\log _{a} b}\)

You can derive this formula by converting \(\ \log_bx\) to exponential form and then taking the log base x of both sides. This is shown below.

\(\ \begin{aligned}
\log _{b} x &=y \\
b^{y} &=x \\
\log _{a} b^{y} &=\log _{a} x \\
y \log _{a} b &=\log _{a} x \\
y &=\frac{\log _{a} x}{\log _{a} b}
\end{aligned}\)

Therefore, \(\ \log _{b} x=\frac{\log _{a} x}{\log _{a} b}\).

If you were to evaluate log34 using your calculator, you may need to use the change of base formula since some calculators only have base 10 or base e . The result would be:

\(\ \log _{3} 4=\frac{\log _{10} 4}{\log _{10} 3}=\frac{\ln 4}{\ln 3} \approx 1.262\)

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Examples

Example 1

Earlier, you were asked how to use a calculator to evaluate an expression like log₃12.

Solution

In order to evaluate an expression like log₃12 you have some options on your calculator:

\(\ \frac{\ln 12}{\ln 3}=\frac{\log 12}{\log 3} \approx 2.26\)

Some graphing calculators also have another option. Press the MATH followed by the A buttons and enter log₃12.

Example 2

Prove the following log identity.

\(\ \log _{a} b=\frac{1}{\log _{b} a}\)

Solution

\(\ \log _{a} b=\frac{\log _{x} b}{\log _{x} a}=\frac{1}{\frac{\log _{x} a}{\log _{x} b}}=\frac{1}{\log _{b} a}\)

Example 3

Simplify to an exact result: \(\ \left(\log _{4} 5\right) \cdot\left(\log _{3} 4\right) \cdot\left(\log _{5} 81\right) \cdot\left(\log _{5} 25\right)\)

Solution

\(\ \frac{\log 5}{\log 4} \cdot \frac{\log 4}{\log 3} \cdot \frac{\log 3^{4}}{\log 5} \cdot \frac{\log 5^{2}}{\log 5}=\frac{\log 5}{\log 4} \cdot \frac{\log 4}{\log 3} \cdot \frac{4 \cdot \log 3}{\log 5} \cdot \frac{2 \cdot \log 5}{\log 5}=4 \cdot 2=8\)

Example 4

Evaluate: \(\ \log _{2} 48-\log _{4} 36\)

Solution

\(\ \begin{aligned}
\log _{2} 48-\log _{4} 36 &=\frac{\log 48}{\log 2}-\frac{\log 36}{\log 4} \\
&=\frac{\log 48}{\log 2}-\frac{\log 6^{2}}{\log 2^{2}} \\
&=\frac{\log 48}{\log 2}-\frac{2 \cdot \log 6}{2 \cdot \log 2} \\
&=\frac{\log 48-\log 6}{\log 2} \\
&=\frac{\log \left(\frac{48}{6}\right)}{\log 2} \\
&=\frac{\log 8}{\log 2} \\
&=\frac{\log 2^{3}}{\log 2} \\
&=\frac{3 \cdot \log 2}{\log 2} \\
&=3
\end{aligned}\)

Example 5

Given log₃5≈1.465 find log₂₅27 without using a log button on the calculator.

Solution

\(\ \log _{25} 27=\frac{\log 3^{3}}{\log 5^{2}}=\frac{3}{2} \cdot \frac{1}{\left(\frac{\log 5}{\log 3}\right)}=\frac{3}{2} \cdot \frac{1}{\log _{3} 5} \approx \frac{3}{2} \cdot \frac{1}{1.465}=1.0239\)

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Review

Evaluate each expression by changing the base and using your calculator.

1. log₆15

2. log₉12

3. log₅25

Evaluate each expression.

4. log₈(log₄(log₃81))

5. log₂3⋅log₃4⋅log₆16⋅log₄6

6. log125⋅log₉4⋅log₄81⋅log₅10

7. log₅(5^log₅125)

8. log(log₆(log₂64))

9. 10^{log₁₀₀9}

10. (log₄x)(log_x16)

11. log₄₉49⁵

12. 3log₂₄24⁸

13. 4^log₂3

Prove the following properties of logarithms.

14. (log_ab)(log_bc)=log_ac

15. (log_ab)(log_bc)(log_cd)=log_ad

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Review (Answers)

To see the Review answers, open this PDF file and look for section 3.5.

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Vocabulary

Term	Definition
Change of Base Formula	Let b, x, and y be positive numbers, b≠1 and y≠1. Then \(\ \log _{y} x=\frac{\log _{b} x}{\log _{b} y}\). More specifically, \(\ \log _{y} x=\frac{\log x}{\log y}\) and \(\ \log _{y} x=\frac{\ln x}{\ln y}\), so that expressions can be evaluated using a calculator.
log	"log" is the shorthand term for 'the logarithm of', as in "logbn" means "the logarithm, base b, of n."
Logarithm	A logarithm is the inverse of an exponential function and is written logba=x such that b^x=a.