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

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


Let $y=\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)$. Then
$$y=\log _{\sqrt{2}}\left((x-2)^{\frac {1}{2}}\right)$$
$$\left( \sqrt {2} \right) ^y = (x-2)^{\frac {1}{2}}$$
$$\left( \left( \sqrt {2} \right) ^y \right) ^2= \left( (x-2)^{\frac {1}{2}} \right) ^2$$
$$\left( \sqrt {2} \right) ^{2y} = x-2$$
$$2^{y} = x-2$$
$$y=\log_2 (x-2)$$
