Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 305: 54

Answer

$ \frac{4}{3}ln(x-4)-\frac{2}{3}ln(x+1)-\frac{2}{3}ln(x-1)$

Work Step by Step

$ln[\frac{(x-4)^2}{x^2-1}]^{2/3}=\frac{2}{3}[ln\frac{(x-4)^2}{(x+1)(x-1)}]=\frac{2}{3}[2ln(x-4)-ln(x+1)-ln(x-1)]=\frac{4}{3}ln(x-4)-\frac{2}{3}ln(x+1)-\frac{2}{3}ln(x-1)$
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