Precalculus (6th Edition) Blitzer

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

Chapter 3 - Review Exercises - Page 513: 57

Answer

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

Work Step by Step

$\displaystyle \frac{1}{2}\ln x-\ln y=\qquad $...apply the Power Rule: $\qquad \log_{b}(M^{p})=p\cdot\log_{b}\mathrm{M}$ $=\ln x^{1/2}-\ln y\qquad $...apply the Quotient Rule: $\displaystyle \qquad \log_{b}(\frac{M}{N})=\log_{b}\mathrm{M}-\log_{b}\mathrm{N}$ $=\displaystyle \ln\frac{x^{1/2}}{y}$
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