Wolfram Alpha gives different answers for doing the same thing to both sides of this equation. I have the equation $10^x=2i$ (I'm doing this because I wanted to take the log of $2i$) I was trying to solve it by hand first, but I was running into errors, so I wanted to make sure when I square both sides of the equation, I wasn't doing anything wrong.
First I entered $10^x=2i$, I got
$$x = \frac {4 i π n + i π + 2 \log2}{2 \log2 + \log5},\ \ \ n\in\mathbb Z $$ 
(log is the natural logarithm in the solution above)
When I did $(10^x)^2=(2i)^2$, it said that is was equivalent to $10^{2x} = -4$ which was to be expected. However, it gave me the following output:
$$x = \frac {2i π n + i π + 2 \log2}{2 \log2 + \log5},\ \ \ n\in\mathbb Z $$ 
Is this a glitch in wolfram alpha? If so, why does this happen? It should give equivalent answers if I did the same thing to both sides of the equations.
If this isn't a glitch, please explain why it isn't a glitch, as I don't understand why this wouldn't be one.

Here is now you would solve your original proposition.
$10^x = e^{x\ln 10}\\
z = x \ln 10 = a+ib\\
e^z = 2i\\
e^{a+ib} = (e^a)(\cos b + i \sin b)\\
e^{a+ib} = 2i\\
e^{a+ib} = 2(\cos (\frac \pi2 + 2n\pi)+i\sin(\frac \pi2 + 2n\pi)\\
a+ib = \ln 2+i(\frac \pi2 + 2n\pi)=x \ln 10\\
x =\frac {\ln 2+i(\frac \pi2 + 2n\pi)}{\ln 10} $
Why does  $10^x = 2i$ have a different solution set from $(10^x)^2 = (2i)^2?$
Because when you square both sides, every solution to $10^x = -2i$ is aslo solution of $(10^x)^2 = (2i)^2$ in addition to the solution that you are looking for.
