## Precalculus (10th Edition)

$\dfrac{f(x+h)-f(x)}{h}=\log_a {(\frac{x+h}{x})^{\frac{1}{h}}}$ Refer to the step-by-step part to see solution.
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, $\dfrac{f(x+h)-f(x)}{h}=\log_a {\left(\frac{x+h}{x}\right)^{\frac{1}{h}}}$