Problem with proof for ClogA=logA^C

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In summary, the logarithm rule states that if y=log_a(x^C), then x^C=a^y. This is proved by induction, using the definition of the exponential function and the chain rule. If m and n are rational, the log rule is proven. If m and n are irrational, then Dedekind cuts must be used to prove the result.
  • #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
 
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  • #2
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].
 
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  • #3
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].

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/1026637So double occurence of same problem
 
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  • #4
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".
 
  • #5
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.

Could you please show this a bit more clearly. Would be much appreciated.
 
  • #6
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".

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?
 

1. What is ClogA=logA^C?

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.

2. What is the significance of ClogA=logA^C?

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.

3. How is ClogA=logA^C used in scientific research?

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.

4. What are some real-life applications of ClogA=logA^C?

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.

5. Are there any limitations to the use of ClogA=logA^C?

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.

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