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: 28

Answer

$ln[16(\frac{x^2+1}{x(x-4)})^{\frac{1}{2}}]$

Work Step by Step

$\frac{1}{2}ln(x^2+1)-4ln\frac{1}{2}-\frac{1}{2}[ln(x-4)+lnx]=ln(x^2+1)^{\frac{1}{2}}+4ln2-ln(x(x-4))^{\frac{1}{2}}=ln[(x^2+1)^{\frac{1}{2}}\times2^4\times(\frac{1}{x(x-4)})^{\frac{1}{2}}]=ln[16(\frac{x^2+1}{x(x-4)})^{\frac{1}{2}}]$
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