College Algebra (11th Edition)

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

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises: 65

Answer

$x=25$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^{3/2}=125 ,$ raise both sides to the exponent equal to $ \dfrac{2}{3} .$ Then use the laws of exponents and the definition of rational exponents to simplify the resulting equation. Finally, do checking if the solution satisfies the original equation. $\bf{\text{Solution Details:}}$ Raising both sides to the exponent, $ \dfrac{2}{3} ,$ the given equation becomes \begin{array}{l}\require{cancel} \left(x^{\frac{3}{2}}\right)^{\frac{2}{3}}=(125)^{\frac{2}{3}} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^{\frac{3}{2}\cdot\frac{2}{3}}=125^{\frac{2}{3}} \\\\ x=125^{\frac{2}{3}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x=\left( \sqrt[3]{125} \right)^2 \\\\ x=\left( 5 \right)^2 \\\\ x=25 .\end{array} Upon checking, $ x=25 $ satisfies the original equation.
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