How do I calculate ln(12) approximate value to two decimals?

Question

(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.

Answer 1

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.

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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.

Answer 2

$\begin{align}\ln(12) &=\ln(10\times1.2)\\ &=\frac{1}{3}\ln(1000)+\ln(1+0.2)\\ &\approx\frac{1}{3}\ln(1024)+0.2-\frac{(0.2)^2}{2}+\frac{(0.2)^3}{3}\\ &\approx\frac{10}{3}\ln(2)+0.2-\frac{0.04}{2}+\frac{0.008}{3}\\ &\approx(3.333)(0.69)+0.2-0.02+0.003\\ &\approx 2.48 \end{align}$

I used the approximation $\ln(2)\approx 0.69$. This needs a separate computation.

Answer 3

We can express $12$ as $$12=\left(\frac32\right)^4\left(\frac43\right)^3$$

So if we can evaluate $\ln\frac32$ and $\ln\frac43$ to sufficient precision, we have $$\ln12=4\ln\frac32+3\ln\frac43$$

It's a good idea to calculate the logarithms to three decimal places, so that the multiplications and additions don't accumulate too much rounding error. Using the series $$\ln(1+x) = x - \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} + \cdots$$ it is enough to calculate $\ln\frac32$ to five terms, and $\ln\frac43$ to four terms.