College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 41

Answer

$x=16$

Work Step by Step

Changing to exponential form, the given expression, $ \log_6(2x+4)=2 $ is equivalent to \begin{array}{l}\require{cancel} 2x+4=6^2 \\\\ 2x+4=36 \\\\ 2x=36-4 \\\\ 2x=32 \\\\ x=\dfrac{32}{2} \\\\ x=16 .\end{array}
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