College Algebra (11th Edition)

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

Chapter 4 - Review Exercises - Page 469: 55

Answer

$x=\frac{\sqrt[4]{10}-7}{2}$

Work Step by Step

$$\log(2x+7)=0.25$$ $$\log(2x+7)=\frac14$$ $$2x+7=10^{\frac14}$$ $$2x+7=\sqrt[4]{10}$$ $$2x=\sqrt[4]{10}-7$$ $$x=\frac{\sqrt[4]{10}-7}{2}$$
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