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Homework Help: On what cases i get 0 denomiator

  1. Feb 13, 2009 #1
    [tex]
    f(x)=\frac{1}{1+ln|x|}
    [/tex]

    the ln|x| part could be -1 when x=e^-1
    correct??
     
  2. jcsd
  3. Feb 13, 2009 #2

    arildno

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    Certainly!

    Do you have any other solutions to the equation 1+ln|x|=0?
     
  4. Feb 13, 2009 #3
    no
    ??
     
  5. Feb 13, 2009 #4

    arildno

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    Well, if ln|x|=-1, what must |x| equal?
     
  6. Feb 13, 2009 #5
    |x|=e^-1
    x=-e^-1
    x=e^-1
     
  7. Feb 13, 2009 #6

    tiny-tim

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    Yes. :smile:
     
  8. Feb 13, 2009 #7
  9. Feb 13, 2009 #8

    HallsofIvy

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    But Arildno's point is that that is not the only value. As you said, the denominator is 0 for [itex]e^{-1}[/itex] or [itex]-e^{-1}[/itex].
     
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