Exponentialgleichungen, Lösungsweg

$8 \cdot 2^{x^2-2x} = 8 \ \ \ | :8 \\ 2^{x^2-2x} = 1 \ \ \ |log_2(\cdot) \\ log_2(2^{x^2-2x}) = log_2(1) \\ x^2-2x = 0 \\ x(x-2)=0 \\ \Rightarrow x_1 = 0 \ ;\ x_2 = 2$

Exponentialgleichungen, Lösungsweg

$\begin{array}{|rcll|} \hline 8\cdot 2^{x^2-2x} &=& 8 \quad & |\quad :8 \\ 2^{x^2-2x} &=& 1 \quad & |\quad 2^0 = 1 \\ 2^{x^2-2x} &=& 2^0 \\ x^2-2x &=& 0 \\ x(x-2) &=& 0 \\\\ \mathbf{ x_1 }&\mathbf{=}& \mathbf{0} \\\\ x_2 - 2 &=& 0 \\ \mathbf{ x_2 }&\mathbf{=}& \mathbf{2} \\ \hline \end{array}$