Logarithm Properties Does $$\log_{a^2}x=\log_a 2x$$
If it does, then how would you prove it?
If it doesn't, then how would I simplify $\log_{a^2} x$ so that it is in the form of $\log_a n$ where $n$ is some polynomial?
 A: We have
$$
\log{_{a^2}}{x}=\frac{\log x}{\log (a^2)}=\frac{\log x}{2\log a}=\frac12\log{_{a}}{x}=\log{_{a}}{(x^{1/2})}\ne \log{_{a}}(2x)
$$ in general.
A: Here's an explanation which doesn't explicitly use the change of base formula:
Let $c = \log_{a^{2}}(x)$ 
Then, from the definition of the $\log$ function:
$(a^{2})^{c} = x $
$\log(a^{2c}) =\log x$
$2c\log a = \log x \Rightarrow c = \frac{\log x}{2\log a}$
Let $d = log_{a}(2x)$
Then $a^{d} = 2x$
$d = \frac{2\log x}{\log a}$
$\Rightarrow d = 4c \Rightarrow d \neq c$
A: Recall the change of base formula, that says: $$\log_{b}a=\dfrac{\log a}{\log b}$$
So,
$$\log_{a^2}x=\dfrac{\log x}{\log a^2}$$
$$\text {But we know, $\log a^2 = 2\log a$, from log properties}$$
$$\log_{a^2}x=\dfrac{\log x}{2\log a}$$

Now, consider the other equation:
$$\log_{a}2x=\dfrac{\log 2x}{\log a}$$
This doesn't look equal..., you can find a counter example pretty quickly. Suppose $x=1$. We have: $\dfrac{\log x}{2\log a} = \dfrac{\log 1}{2\log a} = \dfrac{0}{2\log a} =0$
Other equation, we have:
$\dfrac{\log 2x}{\log a} = \dfrac{\log 2*1}{\log a} = \dfrac{\log 2}{\log a}$
Clearly they are not equal.
