Solve for a (1/3)^(-9x)=a^x

${(\frac{1}{3})}^{- 9 x} = a^{x}$

Rewrite the equation as $a^{x} = {(\frac{1}{3})}^{- 9 x}$ .

$a^{x} = {(\frac{1}{3})}^{- 9 x}$

Simplify ${(\frac{1}{3})}^{- 9 x}$ .

Apply the product rule to $\frac{1}{3}$ .

$a^{x} = \frac{1^{- 9 x}}{3^{- 9 x}}$

One to any power is one.

$a^{x} = \frac{1}{3^{- 9 x}}$