Precalculus (6th Edition) Blitzer

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

Chapter 3 - Cumulative Review Exercises - Page 516: 5


$ x=3$

Work Step by Step

Use the rule $\log (p q)=\log p+\log q $ with $ p=x+5$ and $ q=x-1$ to obtain $\log_{2}(x+5)(x-1)=4$ Simplify: $(x+5)(x-1)=2^4$ $ x^2+4x-5 =16$ or, $(x+7)(x-3)$ So, $ x=-7, 3$, but $ x=-3$ is not in the domain. Thus, our result is: $ x=3$
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