4.2. \( \frac{2^{2 x-1} \cdot 4^{x+1} \cdot 2}{16^{x}} \)

To simplify the expression \( \frac{2^{2 x-1} \cdot 4^{x+1} \cdot 2}{16^{x}} \), we can start by rewriting everything in terms of base 2. 1. Rewrite \( 4^{x+1} \) as \( (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x + 2} \). 2. Rewrite \( 16^{x} \) as \( (2^4)^{x} = 2^{4x} \). Now we can substitute these into the original expression: \[ \frac{2^{2x - 1} \cdot 2^{2x + 2} \cdot 2}{2^{4x}} \] Combine the exponents in the numerator: \[ = \frac{2^{(2x-1) + (2x+2) + 1}}{2^{4x}} = \frac{2^{(2x - 1 + 2x + 2 + 1)}}{2^{4x}} = \frac{2^{4x + 2}}{2^{4x}} \] Now, simplify by subtracting the exponents: \[ = 2^{(4x + 2) - 4x} = 2^2 = 4 \] Thus, the simplified expression is \( 4 \).