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Proof Power Rule Logarithms

  1. Oct 28, 2007 #1
    All information, including the problem, is attached. So far I think I've proven by induction that [tex]log (a^r)[/tex] = [tex]r log (a) [/tex] whenever [tex] r [/tex] is an integer, but I need to prove this for all rational numbers [tex] r = p/q [/tex].

    We're working with the functional equation that has the property that [tex]f(xy) = f(x) + f(y)[/tex], and we're supposed to prove the equality using this. My initial thoughts were to write [tex]f(x*x^{p/q - 1})[/tex] = [tex]f(x) + f(x^{p/q - 1})[/tex], but it didn't get me anywhere. Any thoughts or suggestions?

    Attached Files:

  2. jcsd
  3. Oct 28, 2007 #2


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    [tex]Let log_z b^n =z[/tex]
    and by the definition of logs
    [tex]\frac{z}{n}=log_a b[/tex]
    then multiply by n
  4. Oct 29, 2007 #3


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    You should have said in your post what you say in your attachment: that you are to use the "functional equation" log(xy)= log(x)+ log(y) to prove that log(ar)= r log(a).
    Yes, you can prove that log(an)= n log(a) for any positive integer by induction. Also if n= 0, then a0= 1 so log(a0)= 0= 0log(a).

    If m is a negative integer, then there exist a positive integer n such that m+n= 0.
    log(am+n)= log(an)+ log(am). With m+n= 0, what does that tell you.

    Now, go back and prove that log(anx)= n log(ax) in exactly the same way (or include x from the start) for x any real number. What happens if x= 1/n?
  5. Oct 30, 2007 #4
    Ah I see. Since

    [tex]log (a^{m+n})[/tex] = [tex] log (a^{0})[/tex] = [tex]0[/tex] = [tex] log (a^{n})[/tex] + [tex]log (a^{m})[/tex],

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

    This implies that [tex]nlog(a)[/tex] = [tex](-m)log(a)[/tex], which means that the formula is true for all positive and negative integers, plus zero. Right?

    Now with [tex]r = p/q[/tex], I can write [tex] f(x^{p/q})[/tex] = [tex]f(x^{p*(-q)})[/tex], at which point I can say that because both p and q are integers I have

    [tex]f(x^{p*(-q)})[/tex] = [tex](-p/q)f(x)[/tex].

    Oh wait. I've just realized that [tex]f(x^{-n})[/tex] [tex]\neq[/tex] [tex](-n)f(x)[/tex]. The negative sign shouldn't be there: I should be trying to get

    [tex]f(x^{-n})[/tex] = [tex](1/n)f(x)[/tex].

    Now I'm completely lost. I think I went wrong with the true statement
    [tex]nlog(a)[/tex] = [tex](-m)log(a)[/tex], because I forgot that [tex](-m)[/tex] is positive. After this error I can't seem to get back on track. Help?

    EDIT: Looking at it again, I've realized another mistake:
    [tex]f(x^{p*(-q)})[/tex] = [tex]pf(x)[/tex] - [tex]qf(x)[/tex], not what I stated before.
    Last edited: Oct 30, 2007
  6. Oct 30, 2007 #5


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    Uhhh, no. Assuming, from "am+n= a0", you mean m= -n, this is correct until the last line which should be log(an)= -log(am). That may be just a typo since you have the negative in the next line.


    You mean, of course, f(xp/(-q)) but where did the "-" come from? It's not necessary here. And why did you switch from log to f?

    Don't use both m and n: you mean m+ n= 0 so that m= -n. Just use n and -n.

    Last edited by a moderator: Oct 30, 2007
  7. Oct 30, 2007 #6
    I was mistakenly thinking of [tex]1/x^{-1}[/tex] = [tex]x[/tex] in order to write [tex]p/q[/tex] as [tex]p*(-q)[/tex], but now I see that it should be [tex]p*q^{-1}[/tex], which ruins my entire plan.

    So backtracking, I think now I've reduce the problem to proving that [tex]f(x^{1/q})[/tex] = [tex](1/q)f(x)[/tex], because I already know that I can bring the [tex]p[/tex] down from before so I can ignore it for the moment while I true to prove this property for [tex]1/q[/tex]. Here I'm stuck, though.

    I switched to [tex]f(x)[/tex] because that's how we're supposed to write the problem, and it didn't dawn on me that I should be writing all of the steps of the proof that way until I was half way through.
  8. Oct 30, 2007 #7
    I've thought about your suggestions to go back and prove that

    [tex]log(a^{nx})[/tex] = [tex]n log(a^{x})[/tex] for [tex] x [/tex] as any real number.

    If [tex]x=(1/n)[/tex], then [tex]log(a^{nx})[/tex] = [tex]n log(a^{1/n})[/tex] = [tex] log (a)[/tex].

    [tex]log(a)[/tex] is the same as [tex]n(1/n)log(a)[/tex], but I'm not sure if this proves that

    the [tex]1/n[/tex] can be brought down or if it just shows that in this case such happens to

    be the case.

    Should it prove that that is the case, though, then I will have shown that [tex]1/n[/tex]

    can be treated with the power rule, in which case I can say that [tex]r = p/q[/tex] can

    work with the power rule, which is what I want to prove.
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