what is the equation that will calculate the natural logarithim of a postive number?

ln x = y, where ln is the natural log of number x, is equivalent to
saying e^y = x, where e is an irrational number of approximatly
2.71828. According to mathforum.org [
http://mathforum.org/dr.math/faq/faq.e.html ],

"e is usually defined by the following equation: e = lim n->infinity
(1 + 1/n)^n."

Also from the math forum [
http://mathforum.org/library/drmath/view/51443.html ] is an article
explaining a bit more about natural logs:

"The function ln(x) is the inverse of e^x.  That is, if you flip the
graph of y = e^ x over the line y = x then you would get y = ln(x)."

Also, 

"Now let's say you want to solve ln x = 5:

  you say     e^5 = x

  so            x = 148.413

  so   ln 148.413 = 5"

Search terms:
value for e irrational number

I hope this answers your question. If it does not, please request a
clarification and I will be glad to help with this further.

---

Clarification of Answer by funkywizard-ga on 02 Nov 2002 16:08 PST

thank you sfboy for improving my answer. I'm sorry that I could not
provide quite as good a response as he did, however, I felt that my
answer would have been adaquate. If neither my answer, nor the
comments of sfboy add up to a good answer, I will gladly research the
matter further and add it as a clarification.

Everything in the supposed "answer" is true, but it
doesn't answer the question! In particular, it does
not give a practical way to evaluate ln x for any x.

I presume the aim is to compute ln x in terms of
more standard arithmetic operations such as addition,
multiplication, subtraction, division, and exponentiation.
The Taylor expansion for ln x about the point a = 1
gives:

                   x^2   x^3   x^4   x^5
  ln (x + 1) = x - --- + --- - --- + --- - ...
                    2     3     4     5

Unfortunately, this series converges rather slowly, and
only for values of x near 1. Hence, to compute ln (1.1)
we have:

  ln (1.1) = ln (0.1 + 1) = 0.1 - (0.01)/2 + (0.001)/3 - ...

This converges to ~ 0.09531018 after 8 terms.

For values of x not near 1, you can use the identity:

  ln (a^b) = b * ln a

So, let's say you have a rather large value of x. For
example, imagine x = 100. You can repeatedly take the
square root of x until you get a value close to 1. In
this case:

  sqrt(100) = 100^(1/2) = 10
  sqrt(sqrt(100)) = 100^(1/4) = 3.16227766
  sqrt(sqrt(sqrt(100))) = 100^(1/8) = 1.77827941
  sqrt(sqrt(sqrt(sqrt(100)))) = 100^(1/16) = 1.33352143

Hence, 1.33352143^16 = 100. So:

  ln(100) = ln(1.33352143^16) = 16 * ln(1.33352143).

Plugging x = 0.33352143 into the above equation gives:

  ln(1.33352143) ~ 0.287823135

after 16 terms. Hence,

  ln(100) = 16 * 0.287823135 = 4.60517016

This agrees with the value computed on my calculator
up to the 7th digit after the decimal place. You can
increase the accuracy either by taking more square
roots to get a value closer to 1, or by computing the
series expansion to more terms.

It sure is much easier to do this with the ln x button
on the calculator, though!