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.

It is not WolframAlpha's fault.  To analyze the problem, let us look at a far simpler problem:
$$e^x=-1$$
We all know the solution to this should be given by $x=(2n+1)\pi i$ for $n\in\mathbb Z$.  But upon squaring both sides,
$$e^{2x}=(-1)^2=1$$
The solution to $e^{2x}=1$ is different, as it includes, say, $x=0$, which the original did not.
This is why you want to be careful when manipulating complex numbers :-)
