Problem with proof for ClogA=logA^C

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].

 

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".