## College Algebra (6th Edition)

The solution is $x=9$
$x^{3/2}=27$ Raise both sides of the equation to $\dfrac{2}{3}$: $(x^{3/2})^{2/3}=27^{2/3}$ $x=\sqrt[3]{27^{2}}$ $x=\sqrt[3]{729}$ $x=9$