Proof of logarithmic properties.

AI Thread Summary
The discussion revolves around proving two logarithmic properties: ln(b^(1/n)) = (1/n)ln(b) for b > 0, and ln(a^r) = rln(a) for any rational r and a > 0. The user initially struggles to see the difference between these proofs and previous ones involving natural numbers. It is clarified that the existing proof for natural numbers does not apply to rational numbers because it relies on the multiplication of a base a, which is not feasible for non-integer exponents. The key insight is to express rational numbers as fractions and use the properties of logarithms accordingly. Ultimately, the user gains clarity on how to approach the second proof by leveraging the definition of rational numbers.
pollytree
Messages
14
Reaction score
0

Homework Statement



There are two log properties that I have to prove:
1) Explain why ln(b1/n)=(1/n)ln(b) for b>0, set b=an

2) Explain why ln(ar)=rln(a) for any r in Q and a>0, ie r is rational.

Homework Equations



ln(an)=nln(a)

The Attempt at a Solution



In a previous question I have already proved that ln(an)=nln(a), where n is a natural number. What I'm unsure about, is how is this any different? For 1), I'm not sure why you set b=an? Wouldn't you get ln((an)1/n) = ln(a)? I'm not sure how this helps me find the solution.

Similarly for 2), I'm unsure how it is any different to proving that ln(an)=nln(a).

Any help would be great!
 
Physics news on Phys.org
The obvious difference is that you proved a result m ln(a)=ln(a^m) when m is a natural number, and now they want you to do it when m is not a natural number. The proof that you used for this will not work when m is not a natural number (unless you did something really clever) because at some point you assume that it was, probably in order to write am as a*a*a...*a a multiplied together m times (which doesn't make sense if m=1/2 for example)

If b=an, what does ln(b) and ln(b1/n) become? Use it to do something cool
 
Office_Shredder said:
The obvious difference is that you proved a result m ln(a)=ln(a^m) when m is a natural number, and now they want you to do it when m is not a natural number. The proof that you used for this will not work when m is not a natural number (unless you did something really clever) because at some point you assume that it was, probably in order to write am as a*a*a...*a a multiplied together m times (which doesn't make sense if m=1/2 for example)

If b=an, what does ln(b) and ln(b1/n) become? Use it to do something cool

Thanks for your help. I worked out part 1, but I'm still unsure on part 2. For proving the law for a natural number I did:
Let y=ln(a), for a>0,
ey=a, and if we raise each side to the power n we get:
(ey)n=an
ln(ey*n)=ln(an)
y*n = ln(an), but y = ln(a)
n*ln(a) = ln(an)

So I'm still unsure how it's any different for any rational number?
Any help would be great :)
 
Use the definition of Q, namely that r =m/n, where m,n are in N.
 
Ah I get it now! Thanks :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Replies
9
Views
2K
Replies
12
Views
2K
Replies
13
Views
2K
Replies
6
Views
2K
Replies
13
Views
1K
Replies
3
Views
1K
Replies
10
Views
2K
Back
Top