I am trying to solve $2^x = 3^x$ for $x$.

Now I plug this into WolframAlpha and the steps to the solution are the following:

  • Take natural log of both sides which leaves me with this (according to them):

$$ x\ln(2)= x\ln(3) + 2in\pi $$

What I don't understand is where is this $2in\pi$ coming from?

Thank you very much for your time and help.

  • 3
    $\begingroup$ $e^{x\ln (2)}=e^{x\ln (3)+2in\pi}$ and now consider what $e^{i\theta}$ represents. $\endgroup$ Nov 2 '18 at 16:41
  • 1
    $\begingroup$ wouldn't be easier to divide both sides by $3^x$ and conclude that $x=0$ is the only solution in real numbers? WolframAlpha gives you complex solutions $\endgroup$
    – Vasya
    Nov 2 '18 at 16:41
  • $\begingroup$ The user did tag this question as complex-numbers. $\endgroup$ Nov 2 '18 at 16:47
  • $\begingroup$ Which may very well come from the fact that Wolfram Alpha gave complex numbers as results to the OP. $\endgroup$
    – Ingix
    Nov 3 '18 at 2:25

It results from:


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