# Wolfram Alpha's Why the 2 i π n?

• B
• OmCheeto
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.
OmCheeto
Gold Member
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

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.

Hornbein, topsquark, OmCheeto and 1 other person
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.

FactChecker, topsquark, OmCheeto and 1 other person
Thanks!

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