Precalculus (6th Edition) Blitzer

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

Chapter 6 - Cumulative Review Exercises - Page 801: 18


$log_b(\frac{\sqrt x}{x^2+1})$

Work Step by Step

Using logarithmic properties, we have $\frac{1}{2}log_b(x)-log_b(x^2+1)=log_b(\sqrt x)-log_b(x^2+1)=log_b(\frac{\sqrt x}{x^2+1})$
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