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?

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:

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

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

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

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(5log5125)

8. log(log6(log264))

9. 10log1009

10. (log4x)(logx16)

11. log49495

12. 3log24248

13. 4log23

Prove the following properties of logarithms.

14. (logab)(logbc)=logac

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.