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: 14

Answer

$x\approx4.585$

Work Step by Step

Taking the logarithm of both sides and using the properties of logarithms, the value of the variable that satisfies the given equation, $ 2^x=24 ,$ is \begin{array}{l}\require{cancel} \log2^x=\log24 \\\\ x\log2=\log24 \\\\ x=\dfrac{\log24}{\log2} \\\\ x\approx4.585 .\end{array}
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