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Problem with proof for ClogA=logA^C

  1. Nov 13, 2011 #1

    georg gill

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    hi i have link to problem it is a document from my computer uploaded tp scribd:

    http://www.scribd.com/doc/72599157/PDF-Log-Rule [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 14, 2011 #2

    HallsofIvy

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    How you prove a basic result like that depends upon how you have defined the logarithm! Many texts first define the exponential function, [itex]f(x)= a^x[/itex], prove its propeorties, such as [itex](a^x)^y= a^{xy}[/itex], show that it is "one-to-one" (and so has an inverse), then define [itex]log_a(x)[/itex] as its inverse function. That is, [itex]y= log_a(x)[/itex] if and only if [itex]x= a^y[/itex].

    In particular, if [itex]y= log_a(x^C)[/itex] then [itex]x^C= a^y[/itex]. If [itex]C\ne 0[/itex] taking the "C"th root of both sides, [itex]x= (a^y)^{1/C}= a^{y/C}[/itex]. From that, [itex]y/C= log_a(x)[/itex] so that [itex]y= C log_a(x)[/itex].

    Many modern Calculus texts go the other way- they first define the natural logrithm by
    [tex]ln(x)= \int_1^x \frac{1}{t}dt[/tex].

    From that,
    [tex]ln(x^C)= \int_1^{x^C} \frac{1}{t}dt[/tex]
    If [itex]C\ne 0[/itex] let [itex]u= t^{1/C}[/itex] so that [itex]t= u^C[/itex], [itex]dt= Cu^{C-1}dt[/itex], when t= 1, [itex]u= 1^{1/C}= 1[/itex] and when [itex]t= x^C[/itex], [itex]u= (x^C)^{1/C}= x[/itex] . Then
    [tex]ln(x^C)= \int_1^x \frac{1}{u^C} (Cu^{c-1}du)= C\int_1^x \frac{1}{u}du= Cln(x)[/tex]

    The more general formula follows from the "change of base" formula:
    [tex]log_a(x)= \frac{log_b(x)}{log_b(a)}[/tex]

    [tex]log_a(x^C)= \frac{ln(x^C)}{ln(a)}= C\frac{ln(x)}{ln(a)}= Clog_a(x)[/tex]

    Of course, if C= 0, then [itex]x^C= x^0= 1[/itex] in which case [itex]log_a(x^C)= log_a(1)= 0= C log_a(x)[/itex].
     
    Last edited: Nov 14, 2011
  4. Nov 14, 2011 #3

    georg gill

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    What you write here is wery neat but it is the same problem to me:

    [itex]x= (a^y)^{1/C}= a^{y/C}[/itex]

    I get it if it is a simple problem like:

    [tex](e^2)^4=e^2\cdot e^2\cdot e^2\cdot e^2=e^8[/tex]

    where it only is e to the second power multiplied 4 times and then you multiply e to te second power 4 times and get e multiplied with itself 8 times but if

    [tex](e^2)^{1.32}[/tex]

    then it is a bit harder to imagine and I need a proof to make it look understandable.

    I actually have gotten into a circle of problems here. The log rule above is also proved with derivation

    http://bildr.no/view/1026584
    It uses chain rule which i have proven by linearization and the derivation of lnx is proved in the book by its inverse function

    but in the proof for polynoms they use the same logarithm rule which -I started this with:

    http://bildr.no/view/1026598

    it also assumes that yo know chain rule (as I said no problem) and the derivation of [tex] e^x[/tex] which is proven usin definition of inverse and the fact that the slope of the inverse is 1 divided by the slope of the function it is inverse too. That is dx/dy which make ssense since for the inverse x and y have changed places

    http://bildr.no/view/1026637


    So double occurence of same problem
     
    Last edited: Nov 14, 2011
  5. Nov 14, 2011 #4

    micromass

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    So basically, you want a proof of

    [tex](a^m)^n=a^{mn}[/tex]

    This relies heavily on the definition of the exponentiation.

    If m and n are integers, then it is pretty obvious. A proof by induction should do.

    If m=x/y and n=p/q are fractions, then we recall the definition:

    [tex]a^{x/y}=\sqrt[y]{a^x}[/tex]

    So wee need to prove

    [tex]\sqrt[q]{(\sqrt[y]{a^x})^p}=\sqrt[yq]{a^{xp}}[/tex]

    To prove this, we need some auxiliary results (for m,n integers which are not necessarily the previous m and n)

    1) [itex]\sqrt[m]{a^n}=(\sqrt[m]{a})^n[/itex] This follows basically from the formula

    [tex]\sqrt[m]{ab}=\sqrt[m]{a}\sqrt[m]{b}[/tex]
    Just raise both sides to the m-pth power.

    2) [itex]\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}[/itex]
    Raise both sides to the mn-th power

    With these two results, you can prove the formula. So if m and n are rational, then you have proven it.

    If m and n are irrational, then things are more complicated. I see no other way than actually working with Dedekind cuts to prove the result. You can find the proof in this case in Rudin's "principles of mathematical analysis".
     
  6. Nov 15, 2011 #5

    georg gill

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    Could you please show this a bit more clearly. Would be much appreciated.
     
  7. Nov 27, 2011 #6

    georg gill

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    I think I have prove the things you use in 1):

    http://www.viewdocsonline.com/document/biwlgx

    and

    [tex]\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}[/tex] (I)

    we can write

    [tex]\sqrt[n]{a^m}=\sqrt[n]{aa...a}=y[/tex]

    where (I) is

    [tex]\sqrt[n]{a^m}=\sqrt[n]{a}\sqrt[n]{a}...\sqrt[n]{a}=y[/tex] (I)

    we have [tex]a^{\frac{1}{n}}[/tex] multiplied with itself m times

    and

    [tex]\sqrt[n]{a}=y^{\frac{1}{m}}[/tex] (II)

    from (I) and (II) we get

    [tex](\sqrt[n]{a})^m=y=\sqrt[n]{a^m}[/tex]

    could you please tell the second step one more time?
     
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