Intermediate Algebra (6th Edition)

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

Chapter 9 - Review - Page 598: 130

Answer

$x=-\dfrac{42}{23}$

Work Step by Step

Using the properties of logarithms and since $\log_b x=y$ is equivalent to $b^y=x$, then, \begin{array}{l} \log_6 x-\log_6 (4x+7)=1\\\\ \log_6 \dfrac{x}{4x+7}=1\\\\ \dfrac{x}{4x+7}=6^1\\\\ x=6(4x+7)\\\\ x=24x+42\\\\ -23x=42\\\\ x=-\dfrac{42}{23} .\end{array}
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