Precalculus (6th Edition) Blitzer

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

Chapter 3 - Test - Page 515: 18


No solution.

Work Step by Step

Use the quotient rule $\log_b{\dfrac{p}{q}}=\log_b{p} - \log_b{q}$ with $ p=x-4$ and $ q=x+1$ to obtain $\ln {\dfrac{x-4}{x+1}}=\ln 6$ Simplify to obtain: $\dfrac{x-4}{x+1}=6$ $ x-4 = 6x+6 \implies x=-2$ But we just reject this solution because we can not take the log of a negative number.
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