College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 404: 41


$x=\frac{1}{2}\ln 3\approx 0.5493$

Work Step by Step

We need to solve: $e^{4x}+4e^{2x}-21=0$ We factor: $(e^{2x}+7)(e^{2x}-3)=0$ $(e^{2x}+7)=0$ or $(e^{2x}-3)=0$ $e^{2x}=-7$ or $e^{2x}=3$ $2x=\ln(-7)$ or $2x=\ln 3$ $x=\frac{1}{2}\ln(-7)$=no solution or $x=\frac{1}{2}\ln 3\approx 0.5493$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.