College Algebra (11th Edition)

$x=\{ 0,8 \}$
$\bf{\text{Solution Outline:}}$ 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.