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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]
http://www.scribd.com/doc/72599157/PDF-Log-Rule [Broken]
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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].
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.
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".