Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.3 - Properties of Logarithms - Exercise Set - Page 477: 132


Please see below.

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.