## Precalculus (6th Edition)

$x=\frac{2}{3}$
$\log_{x}\frac{16}{81}=2$ means that x must be raised to the power of 2 to get $\frac{16}{81}$. $x^2=\frac{16}{81}$ $x=\sqrt{\frac{16}{81}}$ $x=\frac{4}{9}$ $x=\frac{2}{3}$