Answer
$\dfrac{f(x+h)-f(x)}{h}=\log_a {(\frac{x+h}{x})^{\frac{1}{h}}}$
Refer to the step-by-step part to see solution.
Work Step by Step
We know that $\log_a {x}-\log_a {y}=\log_a {(\frac{x}{y})}$.
Hence,
\begin{align*}
\dfrac{f(x+h)-f(x)}{h}&=\dfrac{\log_a{(x+h)}-\log_a{x}}{h}\\
&=\frac{1}{h}(\log_a{(x+h)}-\log_a{x})\\
&=\frac{1}{h}\cdot \log_a {\left(\frac{x+h}{x}\right)}
\end{align*}
We know that $\log_a {x^n}=n\cdot \log_a {x}$.
Hence,
$\frac{1}{h}\cdot \log_a {\left(\frac{x+h}{x}\right)}= \log_a {\left(\frac{x+h}{x}\right)^{\frac{1}{h}}}$
Therefore,
$\dfrac{f(x+h)-f(x)}{h}=\log_a {\left(\frac{x+h}{x}\right)^{\frac{1}{h}}}$
