Wolfram Alpha gives different answers for doing the same thing to both sides of this equation.

Question

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.

Answer 1

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 :-)

Answer 2

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.