College Algebra (6th Edition)

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

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 46

Answer

$x\approx0.90$

Work Step by Step

Begin by factoring the terms using trial and error. Then set the terms in each set of parentheses equal to $0$ and solve for $x$ by using the natural log on both sides of each equation. $e^{4x}-3e^{2x}-18=0$ $(e^{2x}+3)(e^{2x}-6)=0$ $e^{2x}=-3, e^{2x}=6$ not possible, $2x=\ln(6)$ $x=\frac{\ln(6)}{2}$ $x\approx0.90$
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