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Finding inverse of a function involving ln.

  1. Sep 28, 2009 #1
    1. Plot f(x)=5+ln(-2x+3), and its inverse.



    2. I know the inverse of ln(x) is exp(1)^(x)., and to find the inverse of a function, solve for x then swap the x and y.



    3. y=5+ln(-2x+3)
    y-5=ln(-2x+3)
    exp(1)^(y-5)=(-2x+3)
    (-1/2)*[exp(1)^(y-5)-3]=x

    y=(-1/2)*[exp(1)^(x-5)-3]

    However when I plot these functions using maple, I notice that they are definately not inverse... :( what am i doing wrong?
     
  2. jcsd
  3. Sep 28, 2009 #2

    CompuChip

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    Looks good (although it beats me why you write exp(1)^p instead of just exp(p) or e^p).
    How can you "see" that they are not inverses?
    If you plug x = (-1/2)*[exp(y-5)-3]=x into 5+ln(-2x+3), you get y right?
     
  4. Sep 28, 2009 #3
    Oh, I guess my theory on the geometry of inverses was a little off. I write exp(1)^p because its the maple command that I was using. Thanks for your comment!
     
  5. Sep 28, 2009 #4

    CompuChip

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    Hehe, you're welcome.
    Sometimes all you need is a little reminder on the definitions :)
     
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