Show that \( \sqrt{2^{x}}+\sqrt{2^{x+4}}=5 \cdot(\sqrt{2})^{x} \) Hence or otherwise solve for \( x \) if \( \sqrt{2^{x}}+\sqrt{2^{x+4}}=2 \frac{1}{2} \)

To solve the equation \( \sqrt{2^{x}}+\sqrt{2^{x+4}}=5 \cdot(\sqrt{2})^{x} \), we can start by rewriting \( \sqrt{2^{x}} \) and \( \sqrt{2^{x+4}} \): \[ \sqrt{2^{x}} = 2^{x/2} \quad \text{and} \quad \sqrt{2^{x+4}} = \sqrt{2^{x} \cdot 2^4} = \sqrt{2^{x}} \cdot 2^{2} = 4\cdot2^{x/2} \] Now substitute these back into the original equation: \[ 2^{x/2} + 4 \cdot 2^{x/2} = 5 \cdot 2^{x/2} \] Combining the left side gives: \[ 5 \cdot 2^{x/2} = 5 \cdot 2^{x/2} \] This shows the initial equation holds true. Now, to solve \( \sqrt{2^{x}}+\sqrt{2^{x+4}}=2 \frac{1}{2} \), we again express in terms of \( 2^{x/2} \): Substituting as before: \[ 2^{x/2} + 4 \cdot 2^{x/2} = 2.5 \Rightarrow 5 \cdot 2^{x/2} = 2.5 \] Dividing both sides by 5 gives: \[ 2^{x/2} = \frac{2.5}{5} = 0.5 \] Now we express \( 0.5 \) in terms of powers of 2: \[ 0.5 = \frac{1}{2} = 2^{-1} \] Thus, we have: \[ 2^{x/2} = 2^{-1} \] Setting the exponents equal gives: \[ \frac{x}{2} = -1 \Rightarrow x = -2 \] Thus, the solution is \( \boxed{-2} \).