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Log(-1)= Real numer

  1. Nov 14, 2007 #1
    Hello guys.

    I have a great issue.

    I need sollution of this problem.

    Log(-1) = Real number.

    Please if you now something about that let me know.
    Please write everything clear and specify all steps.
    Thank you in advance.
  2. jcsd
  3. Nov 14, 2007 #2
    Thank you robert but could you be more specific .
    And how can I apply it in this problem?
  4. Nov 14, 2007 #3
    He is basically saying that ln(-1)=i*pi. However, it is not a real number.
  5. Nov 14, 2007 #4


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    Where's robert?
  6. Nov 14, 2007 #5
    Thank you guys.

    The point is that I already know all the "wrong" answers and I do not need proof that it is an imaginary number.
    This is absolutely clear and defined.
    I need somethink inovative and creative.

    There is a hint about the DEFINITION for Log(-1)

    there should be used some of the logarithmitic properties to be prooved for a real number.

    I am stucked now.

    If somebody can help me please and thank you.
  7. Nov 14, 2007 #6


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    Log(-1) = Log (i^2) = Log(i) + Log (i) = 2 Log(i)
  8. Nov 14, 2007 #7

    Gib Z

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    There is nothing innovative or creative about finding a completely rubbish and wrong answer.
  9. Nov 14, 2007 #8
    That is wrong. [tex]\log (xy)\not =\log (x) +\log(y)[/tex] for [tex]x,y\in \mathbb{C}^{\bold{x}}[/tex].
    Last edited: Nov 15, 2007
  10. Nov 15, 2007 #9


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    log(-1) is not a real number. It simply isn't, by definition of the function.
    But nejnadusho, let us appoint this: if you give me a real solution to [itex]x^2 = -1[/itex] then I'll give you a real expression for log(-1).
  11. Nov 15, 2007 #10
    That's like saying "I need someone to think creatively and help me prove 1+1=3".

    If you could prove any statement you liked, maths would be completely meaningless.
  12. Nov 15, 2007 #11

    Gib Z

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    Unless you only liked correct statements, which all mathematicians should =]
  13. Nov 15, 2007 #12
    Ok guys what if I say like this.

    Log(-1) = x

    => (Log(-1))/(Log10) = x
    => Log(-1) = x*Log10
    => Log(-1) = Log(10^(x)) -where Log10^(x)= 1 therefore
    => Log(-1) = 1

    I know it is crazy but what about that?????
  14. Nov 15, 2007 #13

    Thus endeth this boring discussion.
  15. Nov 15, 2007 #14
    Log(10^(x)) = x
    x*Log10 = x/x
    Log10 = 1
  16. Nov 15, 2007 #15
    The calculation mistake is in this step:

    \textrm{Log}(10^x) = (\textrm{Log}(10))^x

    Use more parentheses to avoid these.
  17. Nov 15, 2007 #16
    ok but

    Log(10^(x)) = 1
    (Log(10))^x = 1^x = 1 always
    Last edited: Nov 15, 2007
  18. Nov 15, 2007 #17
    Your equation


    is wrong, because

    \textrm{Log}(10^x) = x

    and x is not 1 in general.

    At this point I would like to know your age.
  19. Nov 15, 2007 #18


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    nejnadusho, what about this:
    [itex] i^2 = -1 [/itex]
    [itex] (i^2)^2 = (-1)^2 = 1 [/itex]
    [itex] \sqrt{(i^2)^2} = \sqrt{1} [/itex]
    [itex] i^2 = 1 [/itex]
    [itex] i = 1 \implies 0 = 1 - i \implies 0 = \frac{0}{1 - i} = \frac{1 - i}{1 - i} = 1[/itex]
    so 1 = 0 and from this we can prove that log(-1) = 1 (or any real number you like, or anything else you might want to prove but is not true, for that matter).

    And no, there is no error in this calculation, it's just creative thinking.

    I doubt nejnadusho reads any of our posts though, unless they actually conclude that log(-1) is real.
  20. Nov 15, 2007 #19
    opinion about strategy

    CompuChip, your example of mistake is more complicated than the original one given by nejnadusho, and is thus very unlikely going to improve the situation :wink:
  21. Nov 15, 2007 #20


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    Dearly Missed

    OP is just hung up on the mere term "real".
    "Unreal" doesn't exist, or what?

    So, clearly, non-real numbers don't exist. :smile:
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