- #1
georg gill
- 153
- 6
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
http://www.scribd.com/doc/72599157/PDF-Log-Rule
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HallsofIvy said: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].
micromass said: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.
micromass said: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".
ClogA=logA^C is a mathematical equation that relates the logarithm of a number (A) to its base (C). This equation is commonly used in scientific fields such as mathematics, physics, and chemistry.
The equation ClogA=logA^C is significant because it allows us to solve for the unknown value of either A or C when the other variables are known. This can be useful in solving complex equations and analyzing data in various fields of science.
ClogA=logA^C is used in scientific research to solve equations and analyze data in various fields such as biology, chemistry, and physics. It is also used in statistical analysis to determine the relationship between two variables.
ClogA=logA^C has many real-life applications, such as in calculating the pH of a solution in chemistry, determining the rate of decay in radioactive substances, and analyzing data in genetics. It is also used in economics and finance to model and predict growth and decay patterns.
While ClogA=logA^C is a useful equation, it should be noted that it cannot be used for negative values of A or C, and the base (C) must be greater than 0. Additionally, it is important to use caution when applying this equation in certain situations, such as when dealing with very small or large numbers, as it may lead to inaccuracies.