Answer
$ -2ln(x-1) $
Work Step by Step
$ln\frac{x}{x-1}+ln\frac{x+1}{x}-ln(x^2-1)=ln[\frac{x}{x-1}\cdot\frac{x+1}{x}]-ln[(x+1)(x-1)]=ln[\frac{x+1}{x-1}\cdot \frac{1}{(x+1)(x-1)}]=ln\frac{1}{(x-1)^2}=-2ln(x-1), \ x\ne \pm1,0$
Precalculus (10th Edition)
by Sullivan, Michael
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: 63
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