College Algebra (11th Edition)

$x=\{-1024,1024\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $x^{2/5}=16 ,$ raise both sides to the $5$th power to cancel the denominator of the exponent. Then take the square root of both sides and simplify the resulting radical. $\bf{\text{Solution Details:}}$ Raising both sides of the equation to the fifth power, the equation above is equivalent to \begin{array}{l}\require{cancel} \left( x^{\frac{2}{5}} \right)^{5}=(16)^{5} \\\\ x^2=16^{5} .\end{array} Taking the square root of both sides and simplifying the radical, the solutions are \begin{array}{l}\require{cancel} x=\pm\sqrt{16^5} \\\\ x=\pm\left(\sqrt{16}\right)^5 \\\\ x=\pm\left(\sqrt{(4)^2}\right)^5 \\\\ x=\pm\left(4\right)^5 \\\\ x=\pm1024 .\end{array} Hence, the solutions are $x=\{-1024,1024\} .$