Any good approximation methods of $\ln(2)$? If you do a Taylor polynomial for $\ln(x)$ at 1 you can approximate:
$$\ln(2) \approx \sum^n_{k=1} \frac{(-1)^k}{k}$$
The problem is that this converges really slowly, for an error of at most $\frac{1}{n}$ you need to sum $n$ terms.
Are there better approximations?
 A: Another frequently used expansion is
$$
\ln(2)=\ln(\frac43)-\ln(\frac23)=\sum_{k=0}^\infty\frac2{3(2k+1)\cdot9^k}
$$
There are other decompositions with arguments closer to $1$ (similar to the Euler-Machin like formulas for $\pi=4\arctan(1)$), but it is an open question if there is one that gives faster than this kind of linear convergence.
A: Since $\log(2)=-\log\left(\frac12\right)$ and since the Taylor series of the $\log$ function converges fast at $\frac12$, you can use this fact to compute $\log(2)$ quite fast.
A: Maybe apply Newton's method to the equation $e^x -2=0$? The numerical scheme could be something like $x_0=1, \quad x_{n+1} = x_n - \dfrac{e^{x_n}-2}{e^{x_n}}=x_n-1+2e^{-x_n}$. The convergence will be fast, but it implies that you can accurately compute the exponential. The error committed when approximating $\ln 2$ by $x_3$ is close to $0.4 \times 10^{-6}$.
A: This answer adds onto PierreCarre's answer and is too long for a comment.
The cost of multiple evaluations of the exponential function when applying root-finding methods to $e^x-2$ are rapidly decreasing. Note that for Newton's method, one has
$$x_{n+1}=x_n-\Delta x_n,\quad\Delta x_n=1-2\exp(-x_n)$$
so that the next exponential function evaluated is
$$\exp(-x_{n+1})=\exp(-x_n+\Delta x_n)=\exp(-x_n)\exp(\Delta x_n)$$
where $\exp(-x_n)$ is already known and $\exp(\Delta x_n)$ can be computed extremely rapidly. As $x_n$ becomes more accurate, $\Delta x_n$ decreases, so less and less terms are needed to compute $\exp(\Delta x_n)$.
This essentially leaves only the initial evaluation of $\exp(x_0)$, which takes roughly $d/\log_{10}(d)$ iterations to compute to $d$ digits.

Additional notes:
An improved initial estimate should be taken from the other answers, using only their first few terms.
Since higher order derivatives are all just $e^x$, Householder methods such as Halley's method may be easily used to give faster iterative refinement.
