Solve for \( x \). \[ 13^{-2 x}=9^{-x-2} \] Write the exact answer using either base-10 or base-e logarithms \( x=\square \)

To solve the equation \( 13^{-2x} = 9^{-x-2} \), we can start by taking the logarithm of both sides. Taking the natural logarithm (base \( e \)) yields: \[ \ln(13^{-2x}) = \ln(9^{-x-2}) \] Using the property of logarithms \(\ln(a^b) = b \ln(a)\): \[ -2x \ln(13) = (-x - 2) \ln(9) \] Expanding the right side gives: \[ -2x \ln(13) = -x \ln(9) - 2 \ln(9) \] We can rearrange this equation to isolate terms involving \( x \): \[ -2x \ln(13) + x \ln(9) = -2 \ln(9) \] Factoring out \( x \) on the left side: \[ x(-2 \ln(13) + \ln(9)) = -2 \ln(9) \] Thus, solving for \( x \) gives: \[ x = \frac{-2 \ln(9)}{-2 \ln(13) + \ln(9)} \] Simplifying this, we have: \[ x = \frac{2 \ln(9)}{2 \ln(13) - \ln(9)} \] This is the exact solution for \( x \). So, the final answer is: \[ x = \frac{2 \ln(9)}{2 \ln(13) - \ln(9)} \]