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: 46



Work Step by Step

In exponential form, the given logarithmic equation, $ \log_5(7-2x)=3 ,$ is equivalent to \begin{array}{l}\require{cancel} 7-2x=5^3 .\end{array} Hence, the value of the variable that satisfies the given equation is \begin{array}{l}\require{cancel} 7-2x=125 \\\\ -2x=125-7 \\\\ -2x=118 \\\\ x=\dfrac{118}{-2} \\\\ x=-59 .\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.