College Algebra (11th Edition)

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

Chapter 4 - Test - Page 472: 17

Answer

$x=\dfrac{3}{4}$

Work Step by Step

Changing to exponential form, the given equation, $ \log_x \dfrac{9}{16}=2 ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{9}{16}=x^2 .\end{array} Taking the square root of both sides, the solutions of the equation above are \begin{array}{l}\require{cancel} x=\pm\sqrt{\dfrac{9}{16}} \\\\ x=\pm\dfrac{3}{4} .\end{array} Upon checking, only $ x=\dfrac{3}{4} $ satisfies the original equation.
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