College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 178: 32


The solution is $x=9$

Work Step by Step

$x^{3/2}=27$ Raise both sides of the equation to $\dfrac{2}{3}$: $(x^{3/2})^{2/3}=27^{2/3}$ $x=\sqrt[3]{27^{2}}$ $x=\sqrt[3]{729}$ $x=9$
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