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Is log( x^(-y) ) = -y, somebody told me this.

  1. Feb 20, 2005 #1
    Can one under any circumstances say that log x^(-y) = -y ?? I'm having some truble with this myself, but someone told me it is so... :confused:
     
  2. jcsd
  3. Feb 20, 2005 #2

    dextercioby

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    In the complex numbers,your equality (or equation,but i don't know which the unknown is) makes perfect sense...

    Daniel.
     
  4. Feb 20, 2005 #3
    Well, I'm aware of the fact that log(x^(-y))=-y*log(x). But you're saying that what I wrote before makes sense only with complex numbers?
     
  5. Feb 20, 2005 #4

    dextercioby

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    Well,that "-y" has to be greater than 0,in order to make sense among the reals...So "y" should be negative.

    Daniel.
     
  6. Feb 21, 2005 #5

    VietDao29

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    Hi,
    I think that y can be whatever you like (positive, negative, or 0). eg: You can have:
    [tex]\lg{10^{-3}} = -3[/tex]
    So I think:
    [itex]\log_{a}{x^{-y}} = -y[/itex] only if a = x
    Hope I am right,
    Viet Dao,
     
  7. Feb 21, 2005 #6

    dextercioby

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    Yes,apparently my discussion skipped the logarithm part...:mad: Disussed only the exponential.His initial equation is very valid within the reals for x>0 and has the solution,for [itex] y\neq 0 [/itex],[itex] x=e [/itex],the Euler's number.

    Daniel.
     
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