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 - Section 4.5 Properties of Logarithms - 4.5 Assess Your Understanding - Page 331: 56

Answer

$ln5+2ln(x)+\frac{1}{3}ln(1-x)-ln4-2ln(x+1)$

Work Step by Step

$ln\frac{5x^2\sqrt[3] {1-x}}{4(x+1)^2}=ln(5x^2)+ln(\sqrt[3] {1-x})-ln4-ln(x+1)^2=ln5+2ln(x)+\frac{1}{3}ln(1-x)-ln4-2ln(x+1)$
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