The proof of the change of base formula for the logarithms is presented
Let \( y = a^x \) and convert it into logarithm to write
\( \log_a \; y = x \qquad (1) \)
Take the \( \log_b \), where \( b \gt 0 \) and \( b \ne 0 \) , of both sides of \( y = a^x \) to obtain
\( \log_b \; y = \log_b \; (a^x) \)
Use the rules of logarithms to write \( \log_b \; a^x \) as \( x \; \log_b \; a \) and substitute in the above equation.
\( \log_b \; y = x \log_b \; a \)
Substitute \( x \) in the above by \( \log_a \; y \) from \( (1) \) to write
\( \log_b \; y = \log_a \; y \log_b \; a \)
Solve the above for \( \log_a \; y \) to obtain the change of base formula
\[ \log_a \; y = \dfrac{\log_b \; y }{\log_b \; a } \]
You can chose any base \( b \), such that \( b \gt 0 \) and \( b \ne 1 \), to rewrite any logarithm.
a) Evaluate \( \log_4 \; 16 \) noting that \( 16 = 4^2 \)
b) Use the change of base formula to rewrite \( \log_4 \; 16 \) using \( \log \) with base base \( 2 \) and evaluate it again. Compare
Solution
a) \( \log_4 \; 16 = \log_4 \; 4^2 = 2 \)
b) rewrite \( \log_4 \; 16 \) using the change of base formula with logarithm of base \( 2 \),
\( \log_4 \; 16 = \dfrac{\log_2 \; 16 }{\log_2 \; 4 }\)
Note that \( 4 = 2^2 \) and \( 16 = 2^4 \). hence
\( \log_4 \; 16 = \dfrac{\log_2 \; 2^4 }{\log_2 \; 2^2 } = \dfrac{4}{2} = 2\)
The evaluation of \( \log_4 \; 16 \) in a) and b) gives the same answer as expected.
Express \( \log_2 x \) using the natural logarithm \( \ln \)
Solution
Using the change of base formula with the natural logarithm \( \ln \), we write
\( \log_2 x = \dfrac{\ln x}{\ln 2} \)
The results in example 2 have important implications in calculating derivative and integral of logarithms to any base.