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: 36


The solution is $x=\sqrt[5]{27}$

Work Step by Step

$8x^{5/3}-24=0$ Take $24$ to the right side: $8x^{5/3}=24$ Take $8$ to divide the right side: $x^{5/3}=\dfrac{24}{8}$ $x^{5/3}=3$ Raise both sides to $\dfrac{3}{5}$: $(x^{5/3})^{3/5}=3^{3/5}$ $x=\sqrt[5]{3^{3}}$ $x=\sqrt[5]{27}$
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