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

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

 A: $$\log_a(x) = \frac {\log_b(x)} {\log_b(a)}$$ for any $b$.
therefore: $\frac 1 2 \log_{\sqrt 2}(x-2) = \frac {log_2(x - 2)} {2 \log_2(\sqrt 2)} = \frac {log_2(x - 2)} {2 \times \frac 1 2} = \log_2(x-2)$
A: You will need to use the property
$$\log_a (b)=\frac{1}{\log_b (a)}$$
To show that it is true. If you have
$$\frac{1}{2}\log_{\sqrt{2}}(x-2)$$
Then you can change that into
$$\frac{1}{2\log_{x-2}(\sqrt2)}$$
$$\frac{1}{\log_{x-2}(2)}$$
and then flip it again:
$$\log_{2}(x-2)$$
Does that answer your question? If any of my steps were confusing, just let me know.
A: It is a consequence of the change of basis formula 
$$
\log_a x=\frac{\log_b x}{\log_b a}
$$
and of 
$$
\log_2 \sqrt{2}=\frac{1}{2}
$$
A: 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, $$\frac 12 \log_{\sqrt 2} (x-2)=\frac 12 \log_{2^{1/2}} (x-2)=\frac 12 \frac{1}{\frac 12} \log_{ 2} (x-2)= \log_{2} (x-2)$$
A: $\log_2(x-2)$ 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 $$\frac{1}{2}\log_{\sqrt{2}}(x-2) = a = \log_2(x-2).$$
A: 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.
A: Let $y=\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)$. Then
$$y=\log _{\sqrt{2}}\left((x-2)^{\frac {1}{2}}\right)$$
$$\left( \sqrt {2} \right) ^y = (x-2)^{\frac {1}{2}}$$
$$\left( \left( \sqrt {2} \right) ^y \right) ^2= \left( (x-2)^{\frac {1}{2}} \right) ^2$$
$$\left( \sqrt {2} \right) ^{2y} = x-2$$
$$2^{y} = x-2$$
$$y=\log_2 (x-2)$$
