How do I calculate ln(12) approximate value to two decimals? (edit) The problem I've been assigned literally says just to "Calculate ln(12) without a calculator to two decimals using ln(1+x) series." I've been trying to figure out what series I'm dealing with here and don't seem to find an answer anywhere.

When $x> 1/2$ we can use the following series for the logarithm:
$$\ln(x) = \sum_{k = 1}^{+\infty} \dfrac{1}{k} \left(\dfrac{x-1}{x}\right)^k = \dfrac{x-1}{x} + \dfrac{1}{2}\left(\dfrac{x-1}{x}\right)^2 + \dfrac{1}{3} \left(\dfrac{x-1}{x}\right)^3 + \ldots$$
Plugging $x = 12$ you get
$$\ln(12) \approx \dfrac{11}{12} + \dfrac{1}{2} \cdot \dfrac{121}{144} + \dfrac{1}{3}\cdot \dfrac{1331}{1728} = \dfrac{11}{12} + \dfrac{121}{288} + \dfrac{1331}{5184} = \dfrac{8261}{5184} = 1.593(..)$$
Where as $\ln(12) = 2.48$ so you can see this may not be the best way for it converges slowly.

There is another series for the logarithm as pointed out by RobJohn, that works for $x > 0$, that is:
$$\ln(x) = 2\sum_{k = 1}^{+\infty} \dfrac{1}{2k-1} \left(\dfrac{x-1}{x+1}\right)^{2k-1} = 2\left(\dfrac{x-1}{x+1} + \dfrac{1}{3}\left(\dfrac{x-1}{x+1}\right)^3 + \dfrac{1}{5}\left(\dfrac{x-1}{x+1}\right)^5 + \ldots \right)$$
Applying it here gives
$$\ln(12) \approx 2\left(\dfrac{11}{13} + \dfrac{1}{3}\dfrac{11^3}{13^3} + \dfrac{1}{5}\dfrac{11^5}{13^5} + \ldots\right) = \frac{12640826}{5569395} = 2.26(...)$$
Which is rather better.
Add some extra terms to get a two decimal precision. Calculations are easy.
