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### Natural Logarithms

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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/
```
Associated Topics:
High School Logs

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