How can I find the value of $\ln( |x|)$ without using the calculator?

Question

I want to know if there is a way to find for example $\ln(2)$, without using the calculator ?

Thanks

Answer 1

$$\log 2 = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$$ In the general case $$\log \frac{1+x}{1-x} = 2(x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\ldots)$$

Answer 2

The operations that are relatively easy to compute by hand are addition, multiplication, and their inverses, subtraction, and division. With these operations we can compute all rational functions, e.g. $\frac{2x^2-1}{x^3+x-1}$.

We know that $$\ln(x)=\sum_{k=1}^{\infty}(-1)^k\frac{(x-1)^k}{k}$$

for values of $x$ close to $1$. So, if we take partial sums of this series we get approximations to logarithm that only require multiplications and sum and subtractions.

Notice that we only need to be able to compute values of logarithm for numbers close to $1$, since using $\ln(e^kx)=k+\ln(x)$ can allow us to reduce to this case.

Answer 3

And let's not forget this method (read off of the Ln scale).

Image source

Answer 4

How precise do you need the calculation to be?

As a quick and dirty approximation, we know that $2^3 = 8$ and $e^2 \approx 2.7^2 = 7.29$, and so $\ln(2)$ should be just over $\frac{2}{3} \approx 0.67$. Contiuing to match powers, we find $2^{10} = 1024$, and $e^7 \approx (2.7)^7 = (3 - 0.3)^7 = 3^7 -7(3)^6(.3) + 21(3)^5(.3)^2 - 35(3)^4(.3)^3 \dots$ $= 3^7 (1 - .7 + .21 - .035 \dots)$ $\approx 2187(.475) = 1038.825$. Therefore, $e^7 \approx 2^{10}$ and so $\ln(2)$ should be just under $0.7$.

Answer 5

One can use the fact that $$ \log x=\lim_{n\to\infty}n\left(1-\frac{1}{\sqrt[n]{x}}\right) $$ For $\log2$ a good approximation is $$ 1048576\left(1-\frac{1}{\sqrt[1048576]{2}}\right) $$ where $$ \sqrt[1048576]{x} $$ can be computed by pressing twenty times the SQRT key on a pocket calculator, since $1048576=2^{20}$ (or computing it by hand, with much patience and time to spend).

What I get doing those computations is $0.6931469565952$, while a real computer gives $0.69314718055994530941$, so we have five exact decimal digits. Of course bigger numbers won't do, since the $2^{20}$-th root of it will be too near $1$ and the necessary digits would have already been lost.

_{(Note: $\log$ is the natural logarithm; I refuse to denote it in any other way. ;-))}