Logarithm Change of Base Rule
$$ \Large \log_{b}{a} = \dfrac{\log_{c}{a}}{\log_{c}{b}}, \small \text{for }a,b,c>0 \text{ and } b,c \ne 1$$
For example,
\( \begin{align}
\log_{3}{8} &= \dfrac{\log_{2}{8}}{\log_{2}{3}} \\
&= \dfrac{\log_{5}{8}}{\log_{5}{3}} \\
&= \dfrac{\log_{10}{8}}{\log_{10}{3}} \\
&\vdots \\
&= 1.8927 \cdots
\end{align} \)
$\textit{Proof:}$
\( \begin{align} \displaystyle \require{AMSsymbols}
\text{Let } \log_{b}{a} &= x \cdots (1)\\
b^x &= a \\
\log_{c}{b^x} &= \log_{c}{a} &\color{red}{\text{taking logarithm in base }c}\\
x\log_{c}{b} &= \log_{c}{a} \\
x &= \dfrac{\log_{c}{a}}{\log_{c}{b}} \\
\therefore \log_{b}{a} &= \dfrac{\log_{c}{a}}{\log_{c}{b}} &\color{red}{\text{replaced by } (1)}
\end{align} \)
Example 1
Evaluate $\log_{2}{7}$, correct to 3 decimal places.
\( \begin{align} \displaystyle
\log_{2}{7} &= \dfrac{\log_{10}{7}}{\log_{10}{2}} \\
&= 2.80735 \cdots \\
\therefore \log_{2}{7} &= 2.807
\end{align} \)
Example 2
Solve $3^{2x-1}=11$ for $x$, correct to 3 significant figures.
\( \begin{align} \displaystyle
3^{2x-1} &= 11 \\
2x-1 &= \log_{3}{11} \\
2x &= \log_{3}{11} +1 \\
x &= \dfrac{1}{2}(\log_{3}{11} +1) \\
&= \dfrac{1}{2}\left(\frac{\log_{10}{11}}{\log_{10}{3}} +1\right) \\
&= 1.5913 \cdots \\
\therefore x &= 1.59
\end{align} \)
Example 3
Solve $8^x-5(4^x) = 0$ for $x$, correct to 4 significant figures.
\( \begin{align} \displaystyle \require{AMSsymbols} \require{color}
8^x-5(4^x) &= 0 \\
2^{3x}-5(2^{2x}) &= 0 \\
2^{2x}(2^x-5) &= 0 \\
2^x-5 &= 0 &\color{red}2^{2x} \ne 0 \\
2^x &= 5 \\
x &= \log_{2}{5} \\
&= \dfrac{\log_{10}{5}}{\log_{10}{2}} \\
&= 2.3219 \cdots \\
\therefore x &= 2.322
\end{align} \)
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