# How do you evaluate 18^(x-2) = 13^(-2x)?

$x = \frac{2 \log \left(18\right)}{2 \log \left(13\right) + \log \left(18\right)}$
Applying $\log$ to each equation side
$\left(x - 2\right) \log \left(18\right) = - 2 x \log \left(13\right)$
solving for $x$ gives
$x = \frac{2 \log \left(18\right)}{2 \log \left(13\right) + \log \left(18\right)}$