Answer
$x=-\frac{1}{6}$
Work Step by Step
We are given the exponential equation $9^{x}=\frac{1}{\sqrt[3]3}$.
We can express each side using a common base and then solve for $x$.
$9^{x}=(3^{2})^{x}=3^{2x}$
$\frac{1}{\sqrt[3]3}=\frac{1}{3^{\frac{1}{3}}}=3^{-\frac{1}{3}}$
$3^{2x}=3^{-\frac{1}{3}}$
Take the natural log of both sides.
$ln(3^{2x})=ln(3^{-\frac{1}{3}})$
$2xln(3)=-\frac{1}{3}ln(3)$
Divide both sides by $ln(3)$.
$2x=-\frac{1}{3}$
Divide both sides by 2.
$x=-\frac{1}{6}$