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Math Help - Should be a easy problem...

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

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

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Linnus View Post
    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
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  3. #3
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    So I got 2 imaginary numbers for y...

    e^x= a imaginary number...

    How do you ln an imaginary number?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Linnus View Post
    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
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