Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - Chapter Review - Cumulative Review - Page 637: 1

Answer

$x=\pm3$

Work Step by Step

$e^{x^2-9}=1=e^0$. We know that if $a\ne1,-1$, then $a^x=a^y\longrightarrow x=y$, hence here: $x^2-9=0\\x^2=9\\x=\pm3$
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.