# 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}