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Functions + Natural Logs

  1. Dec 2, 2007 #1
    1. The problem statement, all variables and given/known data
    The function f is defined for [tex]x>2[/tex] by [tex] f(x)=\ln x+\ln(x-2)-\ln(x^{2}-4)[/tex]
    a. Express f(x) in the form of [tex](\ln\frac{x}{x+a})[/tex]

    b. Find an expression for [tex]f^{-1}(x)[/tex]
    2. Relevant equations
    ..


    3. The attempt at a solution

    Well, I simplified it to:

    [tex]f(x)=\ln(\frac{x^{2}-2x}{x^{2}-4})[/tex]

    I can't figure what to do, ill keep thinking
     
    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2

    Dick

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    I hope you mean you simplified it to log((x-2)/(x^2-4)). Factor the denominator. What was for dinner?
     
  4. Dec 2, 2007 #3
    Im sorry, I left an ln x out, for some reason I typed \lnx and it appeared as nothing. Should I still factor the denominator? ~ I had chow mein with chicken :D
     
  5. Dec 2, 2007 #4

    Dick

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    I love chicken chow mein. You can still factor the denominator and cancel the x-2 in the numerator.
     
  6. Dec 2, 2007 #5
    Oh I get it!

    So [tex] f(x)=\ln \frac {x(x-2)}{(x+2)(x-2)}[/tex]

    So [tex] f(x)=\ln \frac{x}{x+2} [/tex]

    Awesome, part A solved. Now find the inverse..
    Give me a sec here

    Argh, the only way i know of obtaining inverses is by flipping x and y's.. here it seems a tidbit different. I know that:

    The domain of the inverse is the range of the original function.
    The range of the inverse is the domain of the original function.

    How am I supposed to find the inverse :S, I get that [tex] f^{-1}(x)=(e^{x})(y+2)..[/tex]
     
    Last edited: Dec 2, 2007
  7. Dec 2, 2007 #6
    Any hints?
     
  8. Dec 3, 2007 #7

    HallsofIvy

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    No, a function of "x" cannot have a "y" in it!
    You have to start with [tex]y= ln(\frac{x}{x+2})[/tex]
    The first step is, just as you say, to "swap" x and y:
    [tex]x= ln(\frac{y}{y+2}[/tex]
    Now solve for y. You need to get rid of the log:
    [tex]e^x= \frac{y}{y+2}[/tex]
    Now get rid of the fraction by multiplying both sides by y+ 2.
    [tex]e^x(y+2)= y[/tex]
    This is where you stopped, right? But that "y" on the right is NOT the inverse function because you still haven't solved for y.
    [tex]e^xy + 2e^x= y[/tex]
    Now get y by itself on the left, with no y on the right. Can you do that?
    (How would you solve ay+ b= y for y?)
     
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