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

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

 [/INST] Just do it:
$k \log_a b = \log_a b^k$ no matter what type of base $a$.
(Because $k \log_a b = m \implies \log_a b = m/k \implies a^{m/k} = b \implies a^m = b^k \implies m = \log_a b^k$.)
And $\log_a b = \log_{a^x} b^x = x \log_{a^x} b$ for any valid $x$.
( Because $\log_a b = m \implies a^m = b \implies (a^m)^x =(a^x)^m = b^x \implies \log_{a^x} b^x = x \log_{a^x} b$.)
So ...
$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$
So $\frac 12\log_{\sqrt 2}(x-2) = \frac 12 \log_2 (x-2)^2 = \frac 12*2*\log_2(x-2) = \log_2(x-2)$
Or if that is too slick (I always like to double check things work by fundamental definitions)...
Let $\log_{\sqrt 2} (x-2) = z$ then
$\sqrt{2}^z = (x-2)$ then
$(\sqrt{2}^z)^2 = (x - 2)^2$ then
$2^z = (x-2)^2$ then 
$\log_2 (x-2)^2 = z$
Let $\log_2 (x-2) = w$
The $2^w = (x-2)$
$2^{2w} = (x-2)^2$
$z = \log_2 (x-2)^2 = 2w = 2 \log_2 (x-2)$
So $\frac 12 \log_{\sqrt{2}}(x-2) = \frac 12 z = \frac 12 2w = w = \log_2(x-2)$.
It all works.  Learn and get comfortable with these identities.
</s>