(3) \( \sqrt{\frac{2^{x+3}-2^{x+1}}{2 \cdot 2^{x}}}+1 \)

First, let's simplify the expression you've provided: \[ \sqrt{\frac{2^{x+3}-2^{x+1}}{2 \cdot 2^{x}}}+1 \] We can start by factoring the numerator \( 2^{x+3} - 2^{x+1} \). Factoring out \( 2^{x+1} \), we get: \[ 2^{x+3} - 2^{x+1} = 2^{x+1} (2^2 - 1) = 2^{x+1}(4 - 1) = 2^{x+1} \cdot 3 \] Now we substitute this back into our expression: \[ \frac{2^{x+3}-2^{x+1}}{2 \cdot 2^{x}} = \frac{2^{x+1} \cdot 3}{2 \cdot 2^{x}} = \frac{3 \cdot 2^{x+1}}{2 \cdot 2^{x}} = \frac{3 \cdot 2^{x+1}}{2^{x+1}} = \frac{3}{2} \] Thus, our expression simplifies to: \[ \sqrt{\frac{3}{2}} + 1 \] To further simplify: \[ \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \] So, we have: \[ \frac{\sqrt{6}}{2} + 1 \] This is our simplified expression. Alternatively, we could write \( 1 \) as \( \frac{2}{2} \) to combine the two terms: \[ \frac{\sqrt{6}}{2} + \frac{2}{2} = \frac{\sqrt{6} + 2}{2} \] Therefore, the final result is: \[ \frac{\sqrt{6} + 2}{2} \]