Wolfram Alpha's Why the 2 i π n?

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In summary, Wolfram Alpha provides all solutions to the given equation, including complex solutions, while the expected single solution is x = log(y/a)/b. The extra 2 i π n term in the Wolfram solution accounts for the multiple branches of the logarithm function in the complex plane. The n=0 solution is the same as the expected one, but there are also n complex solutions that satisfy the given requirements.
  • #1
OmCheeto
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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
 
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  • #2
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.
 
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  • #3
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.
 
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  • #4
Thanks!
 
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