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Ln(-x) invalidity?

  1. Jun 1, 2010 #1
    A few weeks ago, during my Further Pure 2 class, we were looking at using hyperbolic functions to solve equations. And we ended up with something along the lines of (e^x-2)(e^x+3) where the roots were Ln(2) and invalid due to Ln(-3) being a nonsensical answer. Now, I quickly reverted to something I had previously learnt. That being when you sqrt a negative number you used to be told it was an invalid answer, until you were taught complex numbers/roots, where this was no longer the case. And my question is, is there possibly a (not necessarily complex/imaginary), but some sort of other solution to the answer of the ln(-x) invalidity? Where like ln(-x) took a value of say like q lots of ln(x) in a similar fashion to how complex roots work. If not, can someone show me or explain why this cannot be? And if perhaps there is an answer to this, is it therefore feasible to suggest that there is no such thing as invalidity, as you can always create some sort of new concept or imaginary value in of which a solution is valid for, where it previously was not?

    Cheers Si.
     
  2. jcsd
  3. Jun 1, 2010 #2
    Re: Ln(-x)

    First understand the notation. Usually when we write ln, we mean the ordinary logarithm of a positive number. However, logarithm also makes sense for negative and complex numbers where the customary notation for that is log even though log in high-school is often used to explicitly mean log-base-10. This multiple use of log often causes problems and one often has to interpret the symbol within the context of the problem: if we're dealing with high-school algegra, then log probably means base-10. If however we're dealing with college Complex Variables, it usually means the complex logarithm base e.

    The complex logarithm is valid for all complex numbers except zero and is most commonly defined in terms of the polar representation of a complex number [itex]x+iy=re^{it}[/itex]. Then:

    [tex]log(re^{i\theta})=\ln(r)+i(\theta+2k\pi)[/tex]

    where [itex]\theta[/itex] is the argument of the complex number and [itex]k=0,\pm 1,\pm2,\cdots[/itex] which makes this a "multi-valued" function although often we consider it's "principal value" with k=0. So in the case of [itex]\log(-x)[/itex] where x for example is 2, then the principal-valued logarithm of -2 is:

    [tex]\log(-2)=\ln(2)+\pi i[/tex]
     
  4. Jun 1, 2010 #3
    Re: Ln(-x)

    Thank you very much, this is the answer I was looking for. I had a feeling that there was going to be a formulation using complex number. Can I now revert you to the last line of my question?
     
  5. Jun 1, 2010 #4
    Re: Ln(-x)

    The zeros to [itex](e^x+3)=0[/itex] are [itex]\log(-3)=\ln(3)+i(\pi+2k\pi)[/itex] and is consistent with the definition of log.

    I can't answer the "always". I would imagine we can't always "invent" a new concept as a solution to something previously considered "invalid".
     
  6. Jun 1, 2010 #5
    Re: Ln(-x)

    Thats interesting. I've always thought that the concept of invalidity was just another way of saying we don't know yet, someones hasn't created a new concept or a new way of thinking to solve this idea. But then I guess every time this occurred, you'd just create a new problem, needing a new solution, which would result in an infinite loop.
     
  7. Jun 1, 2010 #6

    Mentallic

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    Re: Ln(-x)

    What about log(0)? :smile:
     
  8. Jun 1, 2010 #7

    Cyosis

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    Re: Ln(-x)

    Log(0) remains indeterminate.
     
  9. Jun 1, 2010 #8

    Mark44

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    Re: Ln(-x)

    The word you want is "undefined."
     
  10. Jun 1, 2010 #9

    Mentallic

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    Re: Ln(-x)

    I don't believe ln(0) is in indeterminate form, it is definitely undefined. Unless you can give me an example of where ln(0) is indeterminate...

    You've already been shown that this is given a quantity, and we don't have to construct a completely new extension on our number system for log(-1). The complex numbers are enough.
     
  11. Jun 1, 2010 #10

    Cyosis

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    Re: Ln(-x)

    Indeed.
     
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