Should be a easy problem...

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

Thanks

Quote:

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Originally Posted by Linnus

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

Thanks

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it is quadratic in

though you don't have to, let , then you have



now solve for . when you are done, you can solve for

So I got 2 imaginary numbers for y...

e^x= a imaginary number...

How do you ln an imaginary number?

Quote:

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Originally Posted by Linnus

So I got 2 imaginary numbers for y...

e^x= a imaginary number...

How do you ln an imaginary number?

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if you are not doing complex analysis, then just say there is no solution and move on.

otherwise

where and

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