## Intermediate Algebra for College Students (7th Edition)

$x = -\dfrac{1}{6}$
Note that $\dfrac{1}{\sqrt[3]{3}} = \dfrac{1}{3^{\frac{1}{3}}}=3^{-\frac{1}{3}}$. Thus, the given expression is equivalent to: $9^x=3^{-\frac{1}{3}}$ Write $9$ as $3^2$ to obtain: $(3^2)^x=3^{-\frac{1}{3}}$ Use the rule $(a^m)^n=a^{mn}$ to obtain: $3^{2x} = 3^{-\frac{1}{3}}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $2x = -\dfrac{1}{3}$ Divide 2 on both sides of the equation to obtain: $x = -\dfrac{1}{6}$