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: 6


No solution

Work Step by Step

Use the rule $\log (p q)=\log p+\log q $ with $ p=x+4$ and $ q=x+1$ to obtain $\ln[(x+4)(x+1)]=2 \ln (x+3)$ Simplify: $(x+3)^2=(x+4)(x+1)$ $ x^2+6x+9=x^2+5x+4$ or, $ x+9=5x+4 \implies x=-5$ But $ x=-5$ is not in the domain. Thus, our result is: No solution
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