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Rajnish Kumar
science forum beginner
Joined: 07 Jul 2006
Posts: 12
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 Posted: Sat Jul 15, 2006 4:48 am Post subject:
Why is e used as natural log
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Why is e used as natural log when 10 is so convenient |
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José Carlos Santos
science forum Guru
Joined: 25 Mar 2005
Posts: 1111
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| Posted: Sat Jul 15, 2006 6:32 am Post subject:
Re: Why is e used as natural log
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Rajnish Kumar wrote:
| Quote: | Why is e used as natural log when 10 is so convenient
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Because (e^x)' = e^x, whereas (10^x)' = log(10)*10^x.
Best regards,
Jose Carlos Santos |
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Fermat
science forum Guru Wannabe
Joined: 07 Dec 2005
Posts: 119
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| Posted: Sat Jul 15, 2006 8:46 am Post subject:
Re: Why is e used as natural log
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José Carlos Santos wrote:
| Quote: | Rajnish Kumar wrote:
Why is e used as natural log when 10 is so convenient
Because (e^x)' = e^x, whereas (10^x)' = log(10)*10^x.
Best regards,
Jose Carlos Santos
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I was going to say the integral of 1/x is ln(x), by definition. Hence
the inverse is e^x. |
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The Ghost In The Machine
science forum Guru
Joined: 25 Mar 2005
Posts: 1551
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| Posted: Sat Jul 15, 2006 7:00 pm Post subject:
Re: Why is e used as natural log
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On Fri, 14 Jul 2006 21:48:09 -0700, Rajnish Kumar wrote:
| Quote: | Why is e used as natural log when 10 is so convenient
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Why indeed? And yet, e pops up in a fair number of places.
The integral integral(u=1,x) (1/u du), for example, gives a function of
x over the domain (0,+oo), which we'll call log(x). This function has
some interesting properties:
[1] log(1) = 0, as one can see by inspection.
[2] log(1/x) = integral(u=1,1/x) (1/u du)
= integral(v=1,x) (v) (-dv/v^2)
= -integral(v=1,x) (1/v dv)
= -log(x)
[3] log(x*y) = integral(u=1,x*y) (1/u du)
= integral(v=1/x,y) (1/(xv) * x dv)
= integral(v=1/x,y) (1/v dv)
= integral(v=1/x,1) (1/v dv) + integral(v=1,y) (1/v dv)
= -log(1/x) + log(y)
= log(x) + log(y)
[4] log(x^n) = n*log(x) for any integer n; the proof by induction is
almost trivial.
[5] log(x^(1/n)) = (1/n)*log(x) for any integer n.
This function has an inverse, since it's fairly obvious that this function
is strictly monotonic over the realm x = 1, +oo. Call this inverse
exp(y), with domain (-oo, +oo).
Now, for what value of x is it such that log(x) = 1? Or, put another way,
what is exp(1)? x=10 won't cut it, since, if one does an obvious
lower-bound approximation of integral(u=1,10)(1/u du), one gets
1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10 = 4861/2520, which is both greater
than 1 and less than log(10) (by about 0.3736...).
Also, since d(log(x))/dx = 1/x, pretty much by definition,
then if one writes y = exp(x) and wants to compute dy/dx, then
x = log(y) and dy/dx = 1/(dx/dy) = 1/(1/y) = y, or d(exp(x))/dx = exp(x).
(Formalization of this abuse of Leibnitz notation is left to the reader.  )
--
#191, ewill3@earthlink.net
It's still legal to go .sigless. |
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