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I know it's wrong but I don't know WHY

  1. Nov 25, 2004 #1

    Alkatran

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    1/ln(x) = ln(x)^-1 = ln(-x)
    ln(-e) = 1/ln(e) = 1/1 = 1

    ln() is only defined over positive values, but you can find solutions like so... where's the error.
     
  2. jcsd
  3. Nov 25, 2004 #2

    anti_crank

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    ln(x)^-1 = ln(-x) is incorrect. You're probably thinking of ln (1/x) = - ln (x)
     
  4. Nov 25, 2004 #3

    Integral

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    Your trouble is that

    [tex] \ln (-x) \ne (\ln(x))^{-1}) [/tex]
    This is the correct way to do it.
    [tex] \ln (-x) = \ln(-1 * x) = \ln(-1) + \ln(x) [/tex]


    There will be complex solutions to this but none on the Real line.
     
  5. Nov 26, 2004 #4

    Alkatran

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    Thanks, I knew something wasn't right. I should caught on when -x = 1/x :rofl:
     
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