Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Chapter Review - Review Exercises - Page 370: 27


$-2 \ln (x+1)$

Work Step by Step

Use rule $\ln_a{(mn)}=\log_a{m} + \log_a{n}$ to obtain: $\ln [(\dfrac{x-1}{x})(\dfrac{x}{x+1})(\dfrac{1}{x^2-1})] =\ln [\dfrac{1}{(x+1)^2}]$ Now, use $\log_a (\dfrac{m}{n})=\ln m -\ln n$ to obtain: $\ln [\dfrac{1}{(x+1)^2}]=\ln (1)-\ln (x+1)^2$ Finally, use $\log_a{a^m}=m\log_a{m}$ to obtain: $\ln (1)-\ln (x+1)^2=-2 \ln (x+1)$
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