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Thread: natural log question

  1. Oct 22nd 2008, 04:06 AM #1
    gtozoom
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    natural log question

    e^x - 6e^(-x) = 1
    solve for x

    I think you can make 1=lne or e^0 and that helps but i'm still having trouble at the next step.
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  2. Oct 22nd 2008, 04:19 AM #2
    SimonM
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    If you multiply through by e^x you should end up with an equation you can solve
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  3. Oct 22nd 2008, 04:19 AM #3
    Chop Suey
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    Multiply both sides by e^x to get:

    e^{2x} - 6 = e^x

    Rearrange it into a quadratic:

    e^{2x} - e^x - 6 = 0

    Notice that this factors into:

    (e^x-3)(e^x+2)=0

    Can you continue?
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  4. Oct 22nd 2008, 05:01 AM #4
    gtozoom
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    got it, thanks alot guys!
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