College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Summary, Review, and Test - Review Exercises - Page 511: 57

Answer

$= \ln \frac{\sqrt x}{y}$

Work Step by Step

$= \frac{1}{2} \ln x - \ln y$ $= \ln x^{\frac{1}{2}} - \ln y$ $= \ln \sqrt x - \ln y $ $= \ln \frac{\sqrt x}{y}$
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