Natural Logarithms

Date: 11/15/1999 at 19:58:54
From: Michael H.
Subject: Why is a natural logarithm called "natural"

Why is a natural logarithm called "natural"? If you can answer this 
please do so in the simplest possible language. Thank you for any 
help that you can offer.

Sincerely,
Michael Hinojosa

Date: 11/16/1999 at 08:36:27
From: Doctor Jerry
Subject: Re: Why is a natural logarithm called "natural"

Hi Michael,

I think there is no absolutely satisfying answer to this question. 
However, although base 10 logarithms were "natural" at the time, they 
were widely and gratefully used in calculation because we use a base 
10 system. Logarithms are now and then used in a much broader context 
than simplifying the extraction of roots or multiplication and 
division. In that broader context, base e logs, i.e. natural logs, 
make much more sense. I think two reasons can be advanced.

1. In calculus, just as it is much simpler to use radians in 
trigonometric functions, it is much simpler to use natural logs. The 
formulas are simpler.

2. The number e emerges from various contexts in a "natural way," 
whereas 10 never emerges in these contexts. For example, if one thinks 
about the growth in an invested amount over a year given different 
compounding periods, the number e emerges.

If you invest A dollars at r% and your money is compounded n = 1 time 
per year, then at the end of the year you will have

     P1 = A + A*(r/100) = A(1+r/100).

If you invest A dollars at r% and your money is compounded n = 2 times 
per year, then at the end of the year you will have

     P2 = A(1+r/(2*100))^2.

If you invest A dollars at r% and your money is compounded n=3 times 
per year, then at the end of the year you will have

     P3 = A(1+r/(3*100))^3

As n becomes very, very large, what does Pn approach? Answer:

     P = A*e^(r/100).

This is because the expression (1+1/m)^m approaches e = 2.71828... as 
m becomes infinite.

In radioactive decay, the governing equation is dA/dt = k*A, that 
is, the rate of change in the amount A is proportional to the amount 
A remaining. The solution to this differential equation is 
ln(A) = k*t+c, where c is a constant.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/