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Inverse function

  1. Feb 20, 2008 #1
    f(x) = ln(e^2x + e^x + 1)

    I want to find the inverse of this function.

    I get:

    e^y=e^2x + e^x + 1

    If I say then that 1=e^0, then can I say y=2x + x? Therefore, x=y/3?
  2. jcsd
  3. Feb 20, 2008 #2
    You got it right as far as taking the exponential on both sides of the equation. From there it helps to notice that:


    Thus converting the entire right hand side into a quadric equation:


    where [tex]u=\exp(x)[/tex] and whilst treating [tex]\exp(y)[/tex] as a constant, solve this with respect to u
  4. Feb 20, 2008 #3


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    No, you can't. e2x+ ex+ e0 is NOT equal to e2x+ x+ 0

    As Troels said, you can replace ex by u and get the equation ey= u2+ u+ 1 or u2+ u+ (1- ey)= 0 and solve that with the quadratic equation.

    That gives
    [tex]u= \frac{-1\pm\sqrt{4e^y- 3}}{2}[/tex]
    You might think that gives two solutions and so there is no "inverse" function (I did at first) but since u= ex, u cannot be negative: we must take the positive root:
    [tex]u= e^x= \frac{-1+ \sqrt{4e^y- 3}}{2}[/tex]
    [tex]x= ln(\frac{-1+ \sqrt{4e^y- 3}}{2})[/tex]

    and the inverse function is
    [tex]f^{-1}(x)= ln(\frac{-1+ \sqrt{4e^x- 3}}{2})[/tex]
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