Question
Prove that:

$\log _{b^m} a=\frac{1}{m} \log _b a$

Answer

$
\begin{aligned}
\log _{\mathrm{b}^{\mathrm{m}}} \mathrm{a} & =\frac{1}{\mathrm{~m}} \log _{\mathrm{b}} \mathrm{a} \\
\text { L.H.S. } & =\log _{\mathrm{b}^{\mathrm{m}}}^{\mathrm{a}}\\
& =\frac{\log \mathrm{a}}{\log \mathrm{b}^{\mathrm{m}}} \quad \ldots\left[\log _y x=\frac{\log x}{\log y}\right] \\
& =\frac{\log \mathrm{a}}{\mathrm{m} \log \mathrm{b}}=\frac{1}{\mathrm{~m}} \log _{\mathrm{b}} \mathrm{a}=\text { R.H.S. }
\end{aligned}
$

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