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Ln Help

  1. Oct 18, 2005 #1
    I was wondering if anyone could help me with some homework.
    I have the equation y = [(e^x)-3]/[(e^x)+1] and need to find its inverse.
    I take the natural ln of both sides and get :
    lny = (x-3)/(x+1)
    Then I multiple the denominator to the other side and get
    lny(x+1) = x -3
    Then i just add three to both sides and switch the x and y's to get the inverse
    y = lnx (x+1) +3

    My teacher got the answer:
    x = ln [ (3+x)/(1-x)]

  2. jcsd
  3. Oct 18, 2005 #2


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    1.Why do you think [itex]ln(\frac{e^{x}-3}{e^{x}+1})=\frac{x-3}{x+1}[/itex]????

    It is totally wrong, you know.

    2.Also, you have miswritten your teacher's answer.

    3. In order to invert this properly, solve first for [itex]e^{x}[/itex], then take the logarithm.
  4. Oct 18, 2005 #3
    It might be good for you to review some basic properties of logorithms and exponetials at this point, as they are highly important in higher math.
  5. Oct 18, 2005 #4
    My teachers answer is written as :
    inverse function (x) = ln [(3+x)/(1-x)]

    Sorry if I made a mistake in writing it the first time, or maybe the teacher has made an error. Could you show me what I'm supposed to get and how to come to that answer?

    Oh and does the ln[ (e^x)-3 / (e^x) + 1] actually equal:
    x -ln3 / x + ln1
    Last edited: Oct 18, 2005
  6. Oct 18, 2005 #5


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    It is a very bad idea to even try to solve equations like
    [tex]y= \frac{e^x-3}{e^x+1}[/tex] if you don't know the basic "laws of logarithms" and your last question indicates that you don't. As you were told before, first solve for ex.

    [tex]y= \frac{e^x-3}{e^x+1}[/tex] so, multiplying on both sides by ex+ 1,
    [tex] (e^x+1)y= e^x- 3[/tex] or
    [tex] e^xy+ y= e^x- 3[/tex] so
    [tex] e^xy- e^x= -y- 3[/tex]
    [tex]e^x(y- 1)= -y-3[/tex]
    [tex]e^x= \frac{y+3}{1-y}[/tex]
    Finally, [tex]x= ln\left(\frac{y+3}{1-y}\right)[/tex]
    as long as -3< y< 1.
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