B Wolfram Alpha's Why the 2 i π n?

[OmCheeto]

Was going through an old spreadsheet and I re-did some math that I had originally noted that I had done incorrectly. It seemed trivially simple but I wanted to double check with Wolfram to make sure I wasn't missing something.

Here's what I typed in: solve for x when y = a * e^(b*x)

Wolfram Alphas solution: x = (log(y/a) + 2 i π n)/b and a!=0 and y!=0 and b!=0 and n element Z

Why did Wolfram Alpha add the ‘2 i π n’ ?
My solution was x = log(y/a)/b

 

[FactChecker]

Wolfram is giving you all the solutions, including complex solutions. The logarithm function has several branches in the complex plane, each with a valid solution. If you were not expected to know about the complex solutions, then your single solution was the expected one. Otherwise, the Wolfram answer is better.

 

[Ibix]

Plug the answer back in to your formula.$$\begin{eqnarray*}
y&=&ae^{b(\log(y/a)+2i\pi n)/b}\\
&=&ae^{\log(y/a)}e^{2i\pi n}\\
&=&ye^{2i\pi n}
\end{eqnarray*}$$The extra exponential on the right is a complex exponential - it turns out that $e^{i\phi}=\cos(\phi)+i\sin(\phi)$. And that means that $e^{2i\pi n}=1$, so the right hand side in my third line above is also equal to $y$.

As FactChecker says, Wolfram is providing all solutions, including complex ones. The $n=0$ solution is the same as yours and is the only real solution. But $n=\ldots,-3,-2,-1,1,2,3,\ldots$ complex solutions also satisfy the requirements you gave Wolfram.

 

[OmCheeto]

Thanks!