Solve for \( x \). \[ 2^{x-3}=11^{-3 x} \] Round your answer to the nearest thousandth. Do not round any intermediate computations.

To solve the equation \( 2^{x-3} = 11^{-3x} \), we can start by taking the logarithm of both sides. We can use common logarithm (base 10) or natural logarithm (base \( e \)). Here, we'll use natural logarithm (ln). Taking the natural logarithm of both sides, we have: \[ \ln(2^{x-3}) = \ln(11^{-3x}) \] Using the property of logarithms that allows us to bring the exponent in front, we can rewrite this as: \[ (x - 3) \ln(2) = -3x \ln(11) \] Expanding the left side: \[ x \ln(2) - 3 \ln(2) = -3x \ln(11) \] Now we can group the terms involving \( x \): \[ x \ln(2) + 3x \ln(11) = 3 \ln(2) \] Factor out \( x \): \[ x(\ln(2) + 3 \ln(11)) = 3 \ln(2) \] Now, solve for \( x \): \[ x = \frac{3 \ln(2)}{\ln(2) + 3 \ln(11)} \] Next, we calculate the numerical values. Calculating \( \ln(2) \) and \( \ln(11) \): \[ \ln(2) \approx 0.693147 \] \[ \ln(11) \approx 2.397895 \] Now substituting these values into the equation: \[ x = \frac{3 \times 0.693147}{0.693147 + 3 \times 2.397895} \] Calculate the denominator: \[ 0.693147 + 3 \times 2.397895 \approx 0.693147 + 7.193685 = 7.886832 \] Now substituting this back into our expression for \( x \): \[ x \approx \frac{2.079441}{7.886832} \] Calculating \( x \): \[ x \approx 0.2631 \] Rounding to the nearest thousandth, we get: \[ \boxed{0.263} \]