Answer
2
Work Step by Step
We know that if $b\gt0$ and $b\ne1$, then $y=log_{b}x$ is equivalent to $x=b^{y}$ (where $x\gt0$ and $y$ is a real number).
Therefore, $log_{\frac{2}{3}}\frac{4}{9}=2$, because $(\frac{2}{3})^{2}=\frac{2^{2}}{3^{2}}=\frac{4}{9}$.