Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set - Page 825: 12



Work Step by Step

Using the laws of exponents, the given equation, $ 9^x=27 ,$ is equivalent to \begin{array}{l}\require{cancel} (3^2)^x=3^3 \\\\ 3^{2x}=3^3 .\end{array} Since the bases are the same, the exponents can be equated. Hence, \begin{array}{l}\require{cancel} 2x=3 \\\\ x=\dfrac{3}{2} .\end{array}
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.