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 - Page 135: 75


$x=\{ 0,8 \}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^{2/3}=2x^{1/3} ,$ raise both sides to the third power to get rid of the denominator in the fractional exponent. Then express the resulting equation in the form $ax^2+bx+c=0,$ and use the concepts of factoring quadratic equations to solve the resulting equation. Finally, do checking if the solution satisfies the original equation. $\bf{\text{Solution Details:}}$ Raising both sides to the third power, the given equation becomes \begin{array}{l}\require{cancel} \left(x^{2/3}\right)^3=\left( 2x^{1/3} \right)^3 .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^{\frac{2}{3}\cdot3 }=2^3x^{\frac{1}{3}\cdot3} \\\\ x^2=8x \\\\ x^2-8x=0 .\end{array} Factoring the $GCF=x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(x-8)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x-8=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x-8=0 \\\\ x=8 .\end{array} Upon checking, $ x=\{ 0,8 \} $ satisfies the original equation.
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