# How do you solve x^ (-2/3) = 9?

Oct 23, 2017

Write it as $\frac{1}{{\left(\sqrt[3]{x}\right)}^{2}} = 9$
So $\frac{1}{9} = {\left(\sqrt[3]{x}\right)}^{2}$
$\sqrt[3]{x} = \pm \sqrt{\frac{1}{9}} = \pm \frac{1}{3}$
$x = {\left(\pm \frac{1}{3}\right)}^{3} = \pm \frac{1}{27}$