Q.Prove that $\log_{b^m} a=\dfrac{1}{m}\log_b a$.

The logarithm is defined as the operation that undoes exponentiation: for $a>0,a\ne1$, $y=\log_ax \iff a^y=x$ (for $x>0$) — the logarithmic form and exponential form of the same fact, convertible into each other directly by matching base-to-base and exponent-to-log-value. The domain of $\log_ax$ is $(0,\infty)$ and its range is all of $R$; for the natural base $e$, $\log_ex$ is written $\ln x$.

Three laws let logarithms convert products, quotients and powers into sums, differences and multiples: $\log_a(bc)=\log_ab+\log_ac$ (product rule), $\log_a\frac bc=\log_ab-\log_ac$ (quotient rule), and $\log_ab^c=c\log_ab$ (power rule) — repeatedly misapplying these (e.g. writing $\log(x+y)=\log x+\log y$, which is false) is the single most common logarithm error. A logarithm in one base converts to any other base $b$ via the change of base formula $\log_ax=\dfrac{\log_bx}{\log_ba}$; a special case is the reciprocal identity $\log_ax=\dfrac{1}{\log_xa}$. …