# Should be a easy problem...

• Nov 16th 2008, 05:20 PM
Linnus
Should be a easy problem...
Solve for x:
e^(2x)+2e^x+15=0

Thanks
• Nov 16th 2008, 05:22 PM
Jhevon
Quote:

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

Thanks

it is quadratic in $\displaystyle e^x$

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

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

now solve for $\displaystyle y$. when you are done, you can solve for $\displaystyle e^x$
• Nov 16th 2008, 05:32 PM
Linnus
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 PM
Jhevon
Quote:

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

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