# Math Help - Should be a easy problem...

1. ## Should be a easy problem...

Solve for x:
e^(2x)+2e^x+15=0

Thanks

2. Originally Posted by Linnus
Solve for x:
e^(2x)+2e^x+15=0

Thanks
it is quadratic in $e^x$

though you don't have to, let $e^x = y$, then you have

$y^2 + 2y + 15 = 0$

now solve for $y$. when you are done, you can solve for $e^x$

3. So I got 2 imaginary numbers for y...

e^x= a imaginary number...

How do you ln an imaginary number?

4. Originally Posted by Linnus
So I got 2 imaginary numbers for y...

e^x= a imaginary number...

How do you ln an imaginary number?
if you are not doing complex analysis, then just say there is no solution and move on.

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

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

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