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

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


This is a property of logarithms -
$$\color{blue}{\log_{m^a}{n}}=\frac {1}{\log_n m^a}=\frac{1}{a\log_n m}=\color{blue}{\frac {1}{a} \log_{m} n}$$
Therefore, $$\frac 12 \log_{\sqrt 2} (x-2)=\frac 12 \log_{2^{1/2}} (x-2)=\frac 12 \frac{1}{\frac 12} \log_{ 2} (x-2)= \log_{2} (x-2)$$
