Log with $\sqrt x$ base I'd like to know how this simplification happened:
$$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$$

$$
\begin{array}{l}
 \color{red}{2 \log _{2} x+\log _{\frac{1}{2}}(1-\sqrt{x})=\frac{1}{2} \log _{\sqrt{2}}(x-2 \sqrt{x}+2) \quad } \color{blue}{0<x<1} \\
\Leftrightarrow 2 \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2) \\
\Leftrightarrow \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2)-\log _{2} x
\end{array}
$$


$\log_2(x-2)$ is the value $a$ such that $2^a = x-2$.  This means $(\sqrt{2})^{2a} = x-2$, so $\log_{\sqrt{2}}(x-2) = 2a$, or $$\frac{1}{2}\log_{\sqrt{2}}(x-2) = a = \log_2(x-2).$$
