College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Summary, Review, and Test - Cumulative Review Exercises (Chapters 1-4) - Page 514: 4

Answer

$x\approx0.97$

Work Step by Step

Isolate the $e$ term and then use the natural log on both sides. Then solve for $x$. $e^{5x}-32=96$ $e^{5x}=128$ $\ln(e^{5x})=\ln(128)$ $5x=\ln(128)$ $x=\frac{\ln(128)}{5}$ $x\approx0.97$
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