Solve for x \[ \begin{array}{l}8^{2 x-4}=\left(\frac{1}{16}\right)^{x-2} \\ x=\end{array} \]

To solve for \( x \) in the equation \[ 8^{2x-4} = \left(\frac{1}{16}\right)^{x-2}, \] we first rewrite both sides with a common base. We know that \( 8 = 2^3 \) and \( 16 = 2^4 \). Thus, we can rewrite \( \frac{1}{16} \) as \( 16^{-1} = (2^4)^{-1} = 2^{-4} \). Rewriting the left side, we have: \[ (2^3)^{2x-4} = 2^{3(2x-4)} = 2^{6x - 12}. \] For the right side, we get: \[ \left( \frac{1}{16} \right)^{x-2} = (2^{-4})^{x-2} = 2^{-4(x-2)} = 2^{-4x + 8}. \] Now the equation looks like this: \[ 2^{6x - 12} = 2^{-4x + 8}. \] Since the bases are the same, we can set the exponents equal to each other: \[ 6x - 12 = -4x + 8. \] Now, we solve for \( x \) by first adding \( 4x \) to both sides: \[ 6x + 4x - 12 = 8 \implies 10x - 12 = 8. \] Next, adding \( 12 \) to both sides yields: \[ 10x = 20. \] Dividing by \( 10 \): \[ x = 2. \] Therefore, the solution is \[ \boxed{2}. \]