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Dumb question

  1. Jan 27, 2005 #1
    I've been trying to do this for almost an hour now but I'm still not able to show that [1-e^(-x)]/[1-e^(x)]=-e^(-x).

    Thank you.
     
  2. jcsd
  3. Jan 27, 2005 #2

    dextercioby

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    U mean show that
    [tex] \frac{1-e^{-x}}{1-e^{x}}=-e^{-x} [/tex]

    How about cross multiply??

    Daniel.
     
  4. Jan 27, 2005 #3
    No, no. I got [1-e^(-x)]/[1-e^(x)] as an answer to a problem, but in the back of the book the answer is given as -e^(-x). What, I'm asking then, is how to get from my answe to the book's...I know for a fact that they're equal.

    Sorry for the confusion.
     
  5. Jan 27, 2005 #4

    dextercioby

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    How about multiplying both the numerator and the denominator by [itex] e^{x} [/itex] ??

    Daniel.
     
  6. Jan 27, 2005 #5
    It doesn't work. :frown:
     
  7. Jan 27, 2005 #6

    dextercioby

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    It works for me
    [tex] (\frac{1-e^{-x}}{1-e^{x}})(\frac{e^{x}}{e^{x}})=(\frac{e^{x}-1}{1-e^{x}})(\frac{1}{e^{x}})=-1\cdot \frac{1}{e^{x}}=-e^{-x} [/tex]

    Daniel.
     
  8. Jan 27, 2005 #7
    (1 - e^(-x))/(1 - e^x) = (1 - 1/e^x)/(1 - e^x) ...then find a common denominator on the top fraction and you get [(e^x - 1)/e^x]/(1 - e^x) = (e^x - 1)/[(e^x)(1 - e^x)] then -1*(1-e^x)/[(e^x)(1 - e^x)] = -1/e^x = -e^(-x)

    wow i typed that fast, should be right,i included lots of steps, i skipped a few, but included more than what you'd need to write down probably, do you see it now?
     
  9. Jan 27, 2005 #8
    @dextercioby:

    how did you turn (e^x-1)/(1-e^x) into 1? If you multiply it by -1 on the top don't you have to do it to the bottom too?
     
  10. Jan 27, 2005 #9

    dextercioby

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    I didn't...I turned into "-1"...


    Daniel.
     
  11. Jan 27, 2005 #10
    Hello Physicsss,

    multiply your equation [1-e^(-x)]/[1-e^(x)]=-e^(-x)
    by [1-e^(x)].

    You get

    [1-e^(-x)] = -e^(-x) * [1-e^(x)]
    => [1-e^(-x)] = -e^(-x) + (-e^(-x))*(-e^(x))
    =>[1-e^(-x)] = -e^(-x) + 1
    => [1-e^(-x)] = 1-e^(-x)
     
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