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Natural logarithm question

  1. Nov 27, 2009 #1
    Hello,

    I made a mistake in the title of this thread and this question is on general logarithms;

    loga(aloga(x)) = loga(x) ==> aloga(x) = x

    Can someone enlighten me on why loga(aloga(x)) simplifies to loga(x)? Can someone prove why this is true? Futhermore, why does this imply that aloga(x) = x? I am having trouble getting my head around this

    Sincerely,

    Bardagath
     
  2. jcsd
  3. Nov 27, 2009 #2

    Borek

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    Basically you are asking why

    [tex]x = a^{\log_a(x)}[/tex]

    What is log definition?
     
  4. Nov 27, 2009 #3
    Do you know the rules:
    [tex]\log_b(a^x) = x\log_b(a)[/tex]
    [tex]\log_b(b) = 1[/tex]
    If you do, then it follows from first applying the first rules and then seeing:
    [tex]\log_a(x) \log_a(a) = \log_a(x) \times 1 = \log_a(x)[/tex]

    The logarithm is what we call an injective function (also called one-to-one function I believe) which basically means that if two elements a and b are mapped to the same element, i.e. [itex]\log_c(a) = \log_c(b)[/itex], then they must necessarily be the same (a=b) since no two different elements map to the same. Apply this to:
    [tex]\log_a\left(a^{\log_a(x)}\right) = \log_a(x)[/tex]
    if you have already shown that log is injective (otherwise you need to use some other property, but the argument seems to suggest that this is the property used).
     
  5. Nov 27, 2009 #4
    I guess that argument does this. If we know [itex]\log_a(a^x) = x[/itex] for all [itex]x[/itex], we want to use it to show [itex]a^{\log_a x} = x[/itex] for all [itex]x[/itex].
     
  6. Nov 27, 2009 #5

    HallsofIvy

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    There are two ways of approaching this. One is that [itex]log_a(x)[/itex] is defined as the inverse function to [itex]a^x[/itex]. By the definition of "inverse functions", which requires that f(f-1(x))= x and that f-1(f(x))= x, then, [itex]a^{log_a(x)}= x[/itex] and [itex]log_a(a^x)= x[/itex].

    Or, just using the laws of logarithms (which are, after all, derived from the definitions), [itex]log_a(a^x)= x log_a(a)= x[/itex] and, if [itex]y= a^{log_a(x)}[/itex], taking the [itex]log_a[/itex] of both sides, [itex]log_a(y)= log_a(a^{log_a(x)})[/itex][itex]= (log_a(x))(log_a(a))= log_a(x)[/itex] and, since [itex]log_a[/itex] is "one-to-one" function, y= x.
     
    Last edited: Nov 28, 2009
  7. Nov 27, 2009 #6

    Borek

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    Obvious typo - you meant a^, not e^ :)
     
  8. Nov 27, 2009 #7
    I don't know why it didn't fall into place earlier but I woke up today and it fits;

    Yes, loga(x)loga(a) = loga(x) . 1 = loga(x)

    Thankyou very much for your replies!
     
  9. Nov 27, 2009 #8

    HallsofIvy

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    Yes, of course. Thanks. I have edited my post so I can pretend I didn't make that mistake.
     
  10. Nov 28, 2009 #9

    Borek

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    Now your LaTeX is broken :rofl:
     
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