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Log identities

  1. Dec 22, 2004 #1

    quasar987

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    I don't get where these two identities come from:

    [tex](logn)^{logn} = n^{log(logn)}[/tex]

    and

    [tex](logn)^{log(logn)} = e^{(log(logn))^2}[/tex]
     
  2. jcsd
  3. Dec 22, 2004 #2
    I can only think of this roundabout way to show the first:

    log(n)^log(n) = x
    log(log(n)^log(n)) = log(x)
    log(n) * log(log(n)) = log(x)
    10^(log(n) * log(log(n))) = 10^log(x)
    (10^log(n))^log(log(n)) = x
    n^log(log(n)) = x

    So log(n)^log(n) = n^log(log(n)).
     
  4. Dec 22, 2004 #3

    NateTG

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    If you're using the natural logarithm, its usually better to use [itex]\ln[/itex] or [itex]\log_e[/itex] rather than [itex]\log[/itex] which can be interpreted in other ways (for example as [itex]\log_{10}[/itex] or as a log with unspecified base) depending on context.

    The identities are similar:

    [tex]n^{\ln(\ln(n))}=\left(e^{\ln(n)}\right)^{\ln(\ln(n))}=e^{\ln(n) \times \ln(\ln(n))}=e^{\ln(\ln(n)) \times \ln(n)}=\left(e^{\ln(\ln(n))}\right)^{\ln(n)}=\left(\ln(n)\right)^{\ln(n)}[/tex]

    [tex]e^{\left(\ln(\ln(n))\right)^2}=e^{\ln(\ln(n)) \times \ln(\ln(n))}=\left(e ^{\ln(\ln(n))}\right)^{\ln(\ln(n))}=\left(\ln(n)\right)^{\ln(\ln(n))}[/tex]
     
  5. Dec 22, 2004 #4

    quasar987

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    Oh, I see! Well thanks a bunch ! :smile:
     
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