Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.2 Exponential Functions - 4.2 Exercises - Page 432: 82


$\dfrac {1}{2}$

Work Step by Step

$2^{6-3x}=8^{x+1}\Rightarrow 2^{6-3x}=\left( 2^{3}\right) ^{x+1}\Rightarrow 2^{6-3x}=2^{3x+3}\Rightarrow 6-3x=3x+3\Rightarrow 6x=3\Rightarrow x=\dfrac {1}{2}$
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.