## College Algebra (11th Edition)

$x=35$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(x-3)^{2/5}=4 ,$ raise both sides to the exponent equal to $\dfrac{5}{2} .$ Then use the laws of exponents and the concepts of rational exponents to solve the resulting equation. Finally, do checking if the solution satisfies the original equation. $\bf{\text{Solution Details:}}$ Raising both sides to the exponent equal to $\dfrac{5}{2} ,$ the given equation becomes \begin{array}{l}\require{cancel} \left( (x-3)^{2/5} \right)^{5/2}=4^{5/2} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x-3)^{\frac{2}{5}\cdot\frac{5}{2} }=4^{\frac{5}{2}} \\\\ x-3=4^{\frac{5}{2}} .\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-3=\left(\sqrt{4}\right)^{5} \\\\ x-3=\left(2\right)^{5} \\\\ x-3=32 \\\\ x=32+3 \\\\ x=35 .\end{array} Upon checking, $x=35$ satisfies the original equation.