(4) \( \frac{\sqrt{4^{2 n-4}} \cdot 8^{5-2 n}}{\sqrt{16^{6-2 n}}} \)

To simplify the expression \( \frac{\sqrt{4^{2n-4}} \cdot 8^{5-2n}}{\sqrt{16^{6-2n}}} \), we start by simplifying each component. First, rewrite the bases: - \( 4 = 2^2 \) - \( 8 = 2^3 \) - \( 16 = 2^4 \) Now substituting these into the expression: 1. **Simplifying the numerator:** \[ \sqrt{4^{2n-4}} = \sqrt{(2^2)^{2n-4}} = \sqrt{2^{4n-8}} = 2^{(4n-8)/2} = 2^{2n-4} \] \[ 8^{5-2n} = (2^3)^{5-2n} = 2^{3(5-2n)} = 2^{15 - 6n} \] Now combine the two parts of the numerator: \[ \sqrt{4^{2n-4}} \cdot 8^{5-2n} = 2^{2n-4} \cdot 2^{15 - 6n} = 2^{(2n - 4) + (15 - 6n)} = 2^{-4n + 11} \] 2. **Simplifying the denominator:** \[ \sqrt{16^{6-2n}} = \sqrt{(2^4)^{6-2n}} = \sqrt{2^{24 - 8n}} = 2^{(24 - 8n)/2} = 2^{12 - 4n} \] Now we have the simplified expression: \[ \frac{2^{-4n + 11}}{2^{12 - 4n}} = 2^{-4n + 11 - (12 - 4n)} = 2^{-4n + 11 - 12 + 4n} = 2^{-1} = \frac{1}{2} \] Thus, the final answer is: \[ \frac{1}{2} \]