Wikipedia:Reference desk/Archives/Mathematics/2007 November 18

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November 18

Applications

Hi, sorry if this is not the right place to ask as this question is slightly different to those asked so far, but I wondered if anyone knew of a good website (or perhaps there exists a page on wikipedia I didn't find) which gives examples of real life applications and uses of mathematical theorems.

I'm not sure if what I asked was clear or not, but if that's the case just let me know and I'll try to explain better.

Thanks for your help everyone. Ashkelona (talk) 12:58, 18 November 2007 (UTC)

Category:Applied mathematics may be useful, though it doesn't seem to cover mathematical physics, which of course is one of the main areas of application of mathematics. Algebraist 14:34, 18 November 2007 (UTC)

e

From my calculator, I know that $e^{ln(k)}=k$ but I have no idea why. Could someone show me the logic behind it? Thanks asyndeton (talk) 17:44, 18 November 2007 (UTC)

Perhaps because $\ln {(k)}$ is defined to be the number a such that $e^{a}=k$? See e (mathematical constant), logarithm and exponential function. -- Meni Rosenfeld (talk) 18:03, 18 November 2007 (UTC)

I will just add that if you take logs of both sides of the apparent equality, you will find

$ln(k)=ln(k)$

...which implies that the previous equality is precisely an equality. (For the sake of simplicity, I'm assuming $k>0$)

Pallida  Mors 20:25, 18 November 2007 (UTC)

This equality holds true for any logarithm base as well, not just log base e. Let's say log is log base 10, then

$10^{log(k)}=k\,$

and for any log base a where a is greater than 0, then

$a^{log_{a}(k)}=k$.

This equality holds true because how the logarithm is defined. Can you figure out why?—Cronholm¹⁴⁴ 20:43, 18 November 2007 (UTC)

I'm not sure what you're asking, but I think it works because taking the log to base 'a' of 'a' will always be one and so, using log laws, you can 'bring down' the power of $log_{a}(k)$ which will be equal to the other side where you had 'k' and took the log to base 'a' of and so you have equality. asyndeton (talk) 20:54, 18 November 2007 (UTC)

Pallida: You have assumed that $\ln e^{t}=t$ is known, but this is just as fundamental as the original question.

asyndenton: Yes, that's one way to look at it, but is needlessly indirect, since the log laws you mention come after the initial definition, and the equality in question is an immediate corollary of the definition. $\log _{a}k$ is defined to be that number b such that $a^{b}=k$, therefore $a^{\log _{a}k}=k$. -- Meni Rosenfeld (talk) 21:03, 18 November 2007 (UTC)

(ec)From the logarithm article which Meni pointed out (I added the arbitrary a and k), "A logarithm of a given number (k) to a given base (a) is the power to which you need to raise the base (a) in order to get the number (k)." right? So if we were to form this definition into an equality it would be

$a^{log_{a}(k)}=k$

which is precisely the problem you presented, so in fact the answer to your question is that it is the definition of the logarithm is what makes your calculator spit that number out. All the other properties used to prove it true actually follow from this definition.—Cronholm¹⁴⁴ 21:11, 18 November 2007 (UTC)

Yeah, Meni, right. I'm using the fact that $ln$ $e=1$, which is a particular case of your assertion. Just a different way to look at it. Pallida  Mors 00:49, 19 November 2007 (UTC)

Sorry, I stand corrected: Meni's assertion is just the definition of a logarithm. And I am also using the definition throughout my proposition. Pallida  Mors 01:04, 19 November 2007 (UTC)

It's difficult to analyse what exactly we are doing when "proving" something as basic as this. But I find your phrasing problematic. We know that if $a>0,b>0$ then $a=b\iff \ln a=\ln b$ and that $\ln e^{t}=t$, and can therefore conclude that $e^{\ln k}=k\iff \ln e^{\ln k}=\ln k\iff \ln k=\ln k$, so the original statement is true. This is pretty much what "taking logs of both sides" means. But these alleged steps for proving the equality are not more trivial than the equality itself. Just as the definition immediately gives $\ln e^{t}=t$, it also gives $e^{\ln k}=k$. -- Meni Rosenfeld (talk) 12:27, 19 November 2007 (UTC)

Yep. A completely logical concern. Why bother with taking logs and all that stuff when the definition of ln itself suffices? The process I described above is formally right, though (it could of course be made more rigorous). But As Meni says, there's no need for it whatsoever. Pallida  Mors 16:27, 19 November 2007 (UTC)