Solve for x:

e^(2x)+2e^x+15=0

Thanks

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- Nov 16th 2008, 05:20 PMLinnusShould be a easy problem...
Solve for x:

e^(2x)+2e^x+15=0

Thanks - Nov 16th 2008, 05:22 PMJhevon
- Nov 16th 2008, 05:32 PMLinnus
So I got 2 imaginary numbers for y...

e^x= a imaginary number...

How do you ln an imaginary number? - Nov 16th 2008, 05:34 PMJhevon
if you are not doing complex analysis, then just say there is no solution and move on.

otherwise $\displaystyle \ln z = \ln |z| + i \text{Arg}z$

where $\displaystyle z \in \mathbb{C}$ and $\displaystyle - \pi < \text{Arg}z \le \pi$

that gives the principal logarithm, which is usually what they want