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: 48



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. $2^{2x}+2^{x}-12=0$ $(2^{x}+4)(2^{x}-3)=0$ $2^{x}=-4, 2^{x}=3$ not possible, $x=\frac{\ln(3)}{\ln(2)}$ $x\approx1.58$
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.