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 306: 89



Work Step by Step

We know that $\log_a {x}+\log_a {y}=\log_a {(xy)}$, Hence, $\ln{x}+\ln({x+1})+\ln{C}\\ =\ln{[x(x+1)(C)]}\\ =\ln{[Cx(x+1)]}$ Thus, the given equation is equivalent to $\ln{y} = \ln{[Cx(x+1)]}$ Use the rule $\log_a{M}=\log_a{N} \longrightarrow M=N$ to obtain $y=Cx(x+1)$
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