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Natural Logarithm of Negative Numbers

  1. Sep 12, 2007 #1
    Well I came across this when someone asked me this question:

    (-2)^n = 16

    I can clearly see n=4. However, he did this:

    ln((-2)^n) = ln(16)
    n*ln(-2) = ln(16)
    n*ln(2)+n*i*pi = ln(16)

    How can I show that n=4 from this???
  2. jcsd
  3. Sep 12, 2007 #2
    It's because n=4 isn't the only possible solution. Remember that log is a multivalued function (like arcsin) -- so introduce the necessary parametrisation, and see that it can be set to zero.
  4. Sep 12, 2007 #3
    Log is a multivalued function since [tex] e^x [/tex] is a periodic function. Remember that Euler showed that

    [tex] e^{ix} = \cos{x} + i\sin{x} [/tex] and hence we have that [tex] e^x = e^{x + 2\pi i n [/tex] and more general since [tex] a^x = e^{\ln{a} x} [/tex] it's true that [tex] a^x [/tex] is a periodic function.
  5. Sep 12, 2007 #4
    Can someone please Exemplify??? I get what they say; but I am still stuck...
  6. Sep 12, 2007 #5

    D H

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    You can't, because your friend made a mistake. As written, n=4 is not a solution. He should have used

    \ln(-2) &= \text{Ln}(2) + (1 + 2k) \pi i \\
    \ln(16) &= \text{Ln}(16) + 2m\pi i

    where [itex]\text{Ln}(x)[/itex] is the principal value of the natural logarithm and [itex]k[/itex] and [itex]m[/itex] are abitrary integers.

    Applying the above to [itex]n\ln(-2) = \ln(16)[/itex] yields

    [tex]n(\text{Ln}(2) + (1 + 2k) \pi i) = \text{Ln}(16) + 2m\pi i[/tex]

    From this you should be able to show that n=4 is but one of infinitely many solutions and also that n=4 is the only pure real solution.
    Last edited: Sep 12, 2007
  7. Sep 13, 2007 #6
    Thanks for all your input. In the end I see that it is quite a simple problem. However, thanks for putting me on track...
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