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: 45



Work Step by Step

In exponential form, the given logarithmic equation, $ \log_2(8-6x)=5 ,$ is equivalent to \begin{array}{l}\require{cancel} 8-6x=2^5 .\end{array} Hence, the value of the variable that satisfies the given equation is \begin{array}{l}\require{cancel} 8-6x=32 \\\\ -6x=32-8 \\\\ -6x=24 \\\\ x=\dfrac{24}{-6} \\\\ x=-4 .\end{array}
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