Approximation of Natural Logarithm using arithmetic. A friend of mine posed this question to me a couple days ago and it's been bugging me ever since. He told me to take the square root of 5 twenty times, subtract 1 from it, and then multiply it by 2^20, giving me a rough approximation of ln (5). I can't for the life of my figure out why this is.
Also: How can you find the natural log of any positive number X using only arithmetic operations (+, x, -, /) and square root?
 A: Let $a$ be a positive real number.
Then 
$$\lim_{n\to\infty}(a^\frac1n-1)\cdot n=\lim_{n\to\infty}\frac{a^{\frac1n}-1}{\frac1n}=\lim_{n\to\infty} \frac{a^\frac1n\cdot\ln a\cdot\frac{-1}{n^2}}{\frac{-1}{n^2}}=\ln a$$
Your result is a special case of this with $a=5$, and $n=2^{20}$
A: I checked this in Mathematica. It really does seem to converge to Log(5) accurately after many iterations:
a = N[Sqrt[5], 30];
Do[
 a = Sqrt[a], {n, 1, 10000 - 1}]
(a - 1)*2^10000

Output:
1.60943791243410037460075933323
compared to:
N[Log[5], 30]

1.60943791243410037460075933323
A: There is no expression for $\ln$, which is a transcendental function, in terms of radical functions and the usual arithmetic operations, whose combinations are, by contrast, algebraic functions.
You can, however, produce arbitrarily good approximations of $\log$ (or any other continuous function for that matter) on any closed interval $[a, b] \subset (0, \infty)$.
For a relatively simple example, for $x \in (-1, 1)$ we have the Maclaurin series
$$\ln(1 + x) = \sum_{k = 1}^{\infty} (-1)^{k + 1} \frac{x^k}{k},$$
and we can produce variously good approximations to $\ln(1 + x)$ on that interval by simply discarding all of the terms of the series after some point. (Notice that this always produces a polynomial approximation.)
