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Math Help - Equation with 2 solutions

  1. #1
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    Equation with 2 solutions

    e^(2x)-7e^x+12=0 had two solutions

    i got this (7(plus minus)sqrt(7^2-4*e*12))/2e now i replaced e with 2.718 because that is the value of e and I got 1.591 and 0.984. Is this hat I what's suppose to do?
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  2. #2
    Senior Member MaxJasper's Avatar
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    Re: Equation with 2 solutions

    Roots are:
    x1=ln(3)
    x2=ln(4)
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  3. #3
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    Re: Equation with 2 solutions

    Quote Originally Posted by MaxJasper View Post
    Roots are:
    x1=ln(3)
    x2=ln(4)
    How did you get that?
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  4. #4
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Equation with 2 solutions

    Let u=e^x then

    u^2-7u+12=0

    solve for u then plug into u=e^x and find x.

    Attached Thumbnails Attached Thumbnails Equation with 2 solutions-exponential-quadratic.png  
    Last edited by MaxJasper; October 15th 2012 at 08:07 AM.
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  5. #5
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    Re: Equation with 2 solutions

    This is what I got when I solved for u:

    U= (loge^x(-12)-2)/-7

    The log is base e^x

    Am I right?
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  6. #6
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    Re: Equation with 2 solutions

    Hello, RD33!

    \text{Solve for }x\!:\;\;e^{2x}-7e^x+12\:=\:0

    Factor: . (e^x - 3)(e^x - 4) \:=\:0

    Then: . \begin{Bmatrix}e^x-3 \:=\:0 & \Rightarrow & e^x \:=\:3 & \Rightarrow & x \:=\:\ln 3 \\ e^x-4 \:=\:0 & \Rightarrow & e^x \:=\:4 & \Rightarrow & x \:=\:\ln4 \end{Bmatrix}

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