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: 64

Answer

$log\frac{(x+3)(x-1)}{(x+6)(x+1)(x-2)} $

Work Step by Step

Factor the expressions, we have $log[\frac{(x+3)(x-1)}{(x+2)(x-2)}]-log[\frac{(x+6)(x+1)}{x+2}]=log[\frac{(x+3)(x-1)}{(x+2)(x-2)}\cdot \frac{x+2}{(x+6)(x+1)}]=log\frac{(x+3)(x-1)}{(x+6)(x+1)(x-2)} $
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