Basic Logarithm Equation $\log_2(x) = \log_x(2) $ 

Using the change of base theorem: $\dfrac{\log(x)}{\log(2)} = \dfrac{\log(2)}{\log(x)}$

Multiplied the denominators on both sides: $\log(x)\log(x) = \log(2)\log(2)$

I kind of get stuck here. I know that you can't take the square root of both sides of the equation, but still, $x = 2$ seems to be an obvious solution to the equation. I've missed $2^{-1}$ or $\frac12$ as another answer to the equation, which I am struggling to get to. 

Any help will be greatly appreciated, thanks in advance.
You have it, actually. 
\begin{align}
(\log(x))^2 &amp;= (\log(2))^2 \\
\log(x) &amp;= \pm \log(2)
\end{align}

For the "$+$" case, you've already solved it.

In the "$-$" case, you have $\log(x) = -\log(2) = \log(2^{-1}) = \log(\frac12)$, from which you can get $x=\frac12$.
log()log()=log(2)log(2)--> log(x)=±log(2)
let a=log(x), a=±log(2), $10^{a}=x$, $x=10^{±log(2)}$.
$x=10^{log(2)}=2$, or $x=10^{-log(2)}$=$(10^{log(2)})^{-1}$=$2^{-1}$=0.5

x=2 or 0.5