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Math Help - Finding the inverse of a function

  1. #1
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    Finding the inverse of a function

    Hi. For my homework one of the first things I'm supposed to be able to do is find the inverse of

    y = (4+e^x)/(4-e^(-x))

    I don't get how this is possible. When I try to find the natural log of both sides I end up with

    lny = ln(4+e^x) - ln(4-e^x)

    There must be some algebra I'm not getting here, because I know there's no log rule that can help me get that X out of the ln

    Thanks, the homework is due tomorrow
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  2. #2
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    Quote Originally Posted by chubigans View Post
    Hi. For my homework one of the first things I'm supposed to be able to do is find the inverse of

    y = (4+e^x)/(4-e^(-x))

    I don't get how this is possible. When I try to find the natural log of both sides I end up with

    lny = ln(4+e^x) - ln(4-e^x)

    There must be some algebra I'm not getting here, because I know there's no log rule that can help me get that X out of the ln

    Thanks, the homework is due tomorrow
    Solve the equation y=\dfrac{4+e^x}{4-e^{-x}} for x. Afterwards swap the variables:

    y=\dfrac{4+e^x}{4-e^{-x}}~\implies~4y-ye^{-x}=4+e^x Multiply through by e^x

    4ye^x-y=4e^x+e^{2x}~\implies~e^{2x} +4e^x(1-y)+y=0

    This is a quadratic equation in e^x:

    e^x=-2(1-y)\pm\sqrt{4(1-y)^2-y}

    Therefore:

    x=\ln\left( 2y-2 + \sqrt{4y^2-9y+4}  \right)~\vee~x=\ln\left( 2y-2 - \sqrt{4y^2-9y+4}  \right)

    Now swap the variables and you'll get the inverse of the function.
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  3. #3
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    Just wanted to say thank you, I can't believe I didn't see that!
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