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3.4.4: Change of Base

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  • 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}

    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:

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


    Example 1

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


    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. 

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


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

    Example 3

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


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

    Example 4

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


    \(\ \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} \\


    Example 5

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


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


    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

    15. (logab)(logbc)(logcd)=logad

    Review (Answers)

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


    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 bx=a.
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