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

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

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  • $\begingroup$ Why don't you check one of the (good) answers given, like that of @Emilio Novati ? $\endgroup$
    – Jean Marie
    Jun 22 '17 at 22:03
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$$\log_a(x) = \frac {\log_b(x)} {\log_b(a)}$$ for any $b$.

therefore: $\frac 1 2 \log_{\sqrt 2}(x-2) = \frac {log_2(x - 2)} {2 \log_2(\sqrt 2)} = \frac {log_2(x - 2)} {2 \times \frac 1 2} = \log_2(x-2)$

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    $\begingroup$ For any b --- $b>0 \, b \neq 1$ $\endgroup$ Jun 21 '17 at 20:05
  • $\begingroup$ Why of course. Implicitly I meant for every $b$ for which the $log_b$ function is defined $\endgroup$
    – AsafHaas
    Jun 21 '17 at 20:07
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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.

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It is a consequence of the change of basis formula $$ \log_a x=\frac{\log_b x}{\log_b a} $$

and of $$ \log_2 \sqrt{2}=\frac{1}{2} $$

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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, $$\frac 12 \log_{\sqrt 2} (x-2)=\frac 12 \log_{2^{1/2}} (x-2)=\frac 12 \frac{1}{\frac 12} \log_{ 2} (x-2)= \log_{2} (x-2)$$

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$\log_2(x-2)$ 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 $$\frac{1}{2}\log_{\sqrt{2}}(x-2) = a = \log_2(x-2).$$

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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.

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

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