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

To solve the equation \( 1.1 \cdot 2 \sqrt{2} \cdot 2^{x + 1} = (0.5)^{x} \), first notice that \( 0.5 \) can be rewritten as \( 2^{-1} \). Thus, the right side becomes \( (2^{-1})^{x} = 2^{-x} \). This transforms the equation into \( 1.1 \cdot 2 \sqrt{2} \cdot 2^{x + 1} = 2^{-x} \). Now, simplify the left side: \( 2^{x + 1} = 2^x \cdot 2 \), giving us \( 1.1 \cdot 2 \sqrt{2} \cdot 2^{x} \cdot 2 = 2^{-x} \), which simplifies to \( 2^{x + 2} \) on the left side. Therefore, we end up with \( 1.1 \cdot 2 \sqrt{2} \cdot 2^{x + 2} = 2^{-x} \). Bringing all factors involving \( x \) on one side, we balance the equation by taking logarithms or equating exponents. Fun fact: solving such exponential equations often involves transforming and simplifying bases, so keep a sharp mind! To verify your solution, substitute values back into the original equation. Mistakes often happen when manipulating properties of exponents, so double-checking your steps can prevent mix-ups in powers. Happy solving!