Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review - Page 670: 27d



Work Step by Step

Let $2^{\log_2 6}=x$. Taking the logarithm base $2$ of both sides, then, \begin{array}{l} \log_2 2^{\log_2 6}=\log_2x \\ (\log_2 6)(\log_2 2)=\log_2x \\ (\log_2 6)(1)=\log_2x \\ \log_2 6=\log_2x .\end{array} Since the bases on both sides of the equal sign are the same, then the logarithm base $2$ can be dropped, resulting to $ x=6 .$
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