implify the following \( \frac{4^{x} \cdot 2^{x-2}}{8^{x-1}} \)

First, let's re-write the expression in terms of powers of 2. We know that \( 4 = 2^2 \) and \( 8 = 2^3 \). So we can express everything as follows: \[ 4^x = (2^2)^x = 2^{2x} \] \[ 8^{x-1} = (2^3)^{x-1} = 2^{3(x-1)} = 2^{3x - 3} \] Now, substitute these into the original expression: \[ \frac{4^{x} \cdot 2^{x-2}}{8^{x-1}} = \frac{2^{2x} \cdot 2^{x-2}}{2^{3x-3}} \] Combine the powers in the numerator: \[ 2^{2x} \cdot 2^{x-2} = 2^{2x + x - 2} = 2^{3x - 2} \] So now the expression becomes: \[ \frac{2^{3x - 2}}{2^{3x - 3}} = 2^{(3x - 2) - (3x - 3)} = 2^{1} = 2 \] Thus, the simplified form of the original expression is: \[ \boxed{2} \]