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

To simplify the expression \( \frac{\sqrt{4^{2 n-4}} \cdot 8^{5-2 n}}{\sqrt{16-2 n}} \), let's break it down step-by-step. 1. **Simplify the terms in the numerator:** - \(\sqrt{4^{2 n-4}} = 4^{(2 n - 4)/2} = 4^{n - 2} = (2^2)^{n - 2} = 2^{2(n - 2)} = 2^{2n - 4}\). - Now for \(8^{5-2n}\), we know that \(8 = 2^3\), so: \[ 8^{5-2n} = (2^3)^{5-2n} = 2^{3(5 - 2n)} = 2^{15 - 6n}. \] - Thus, combining these, we have: \[ \sqrt{4^{2 n-4}} \cdot 8^{5-2 n} = 2^{2n - 4} \cdot 2^{15 - 6n} = 2^{(2n - 4) + (15 - 6n)} = 2^{-4n + 11}. \] 2. **Simplify the denominator:** - The denominator \(\sqrt{16 - 2n}\) can be expressed as: \[ \sqrt{16 - 2n} = \sqrt{2^4 - 2n}. \] 3. **Combine the simplified parts:** - Now the entire expression looks like this: \[ \frac{2^{-4n + 11}}{\sqrt{16 - 2n}}. \] This is the simplified version of the expression. If a further simplification of the denominator is possible (like factoring or another expression), you'd need the value of \(n\) for specific cases, but that's a clean reduction of the original expression.