Logarithm Change of Base Rule

$$\log_{b}{a} = \dfrac{\log_{c}{a}}{\log_{c}{b}}$$ $$\text{for }a,b,c>0 \text{ and } b,c \ne 1$$
For example,
\( \begin{align} \log_{3}{8} &= \dfrac{\log_{2}{8}}{\log_{2}{3}} \\ &= \dfrac{\log_{5}{8}}{\log_{5}{3}} \\ &= \dfrac{\log_{10}{8}}{\log_{10}{3}} \\ &\vdots \\ &= 1.8927 \cdots \\ \end{align} \)

$\textit{Proof:}$

\( \begin{align} \displaystyle \text{Let } \log_{b}{a} &= x \cdots (1)\\ b^x &= a \\ \log_{c}{b^x} &= \log_{c}{a} &\text{taking logarithm in base }c\\ x\log_{c}{b} &= \log_{c}{a} \\ x &= \dfrac{\log_{c}{a}}{\log_{c}{b}} \\ \therefore \log_{b}{a} &= \dfrac{\log_{c}{a}}{\log_{c}{b}} &\text{replaced by } (1)\\ \end{align} \)

Example 1

Evaluate $\log_{2}{7}$, correct to 3 decimal places.

Example 2

Solve $3^{2x-1}=11$ for $x$, correct to 3 significant figures.

Example 3

Solve $8^x - 5(4^x) = 0$ for $x$, correct to 4 significant figures.