## Precalculus (6th Edition) Blitzer

Assuming $f(x)=\log_bx$ and $h\neq 0$, by applying the quotient and power rule, we have$$\frac{f(x+h)-f(x)}{h}=\frac{\log_b (x+h)-\log_bx}{h}=\frac{\log_b \left ( \frac{x+h}{x} \right )}{h}=\frac{1}{h} \log_b \left ( 1+ \frac{h}{x} \right )= \log_b \left ( 1+ \frac{h}{x} \right )^{\frac{1}{h}}$$