\( 1.1 .2 \quad \sqrt{2} \cdot 2^{x+1}=(0,5)^{x} \)

To solve the equation \( \sqrt{2} \cdot 2^{x+1} = (0.5)^{x} \), first recognize that \( (0.5)^{x} = (2^{-1})^{x} = 2^{-x} \). This allows us to rewrite the equation as: \[ \sqrt{2} \cdot 2^{x+1} = 2^{-x} \] Now simplify the left side: \[ \sqrt{2} \cdot 2^{x+1} = \sqrt{2} \cdot 2^x \cdot 2 = 2\sqrt{2} \cdot 2^{x} \] So we have: \[ 2\sqrt{2} \cdot 2^{x} = 2^{-x} \] Dividing both sides by \( 2^{x} \) (assuming \( x \neq 0 \), which we will check later): \[ 2\sqrt{2} = 2^{-2x} \] Taking the logarithm (base 2) of both sides gives: \[ \log_2(2\sqrt{2}) = -2x \] Calculating \( \log_2(2\sqrt{2}) \): \[ \log_2(2\sqrt{2}) = 1 + \log_2(\sqrt{2}) = 1 + \frac{1}{2} = \frac{3}{2} \] Thus: \[ -\frac{3}{2} = -2x \implies 2x = \frac{3}{2} \implies x = \frac{3}{4} \] We should verify if \( x = 0 \) is a valid solution. Plugging \( x = 0 \) back into the original equation: \[ \sqrt{2} \cdot 2^{0+1} = \sqrt{2} \cdot 2^1 = 2\sqrt{2} \] \[ (0.5)^{0} = 1 \] Since \( 2\sqrt{2} \neq 1 \), \( x = 0 \) is not a solution. Therefore, the final solution is: \[ x = \frac{3}{4} \]