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


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    $\begingroup$ Yes:) Look it up in a logarithm table. $\endgroup$ – gammatester Oct 18 '13 at 12:28
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    $\begingroup$ $$\frac{1-2^{1-1.1}}{1.1-1}=0.66967$$ $$\frac{1-2^{1-1.01}}{1.01-1}=0.69075$$ $$\frac{1-2^{1-1.001}}{1.001-1}=0.692907$$ $$\frac{1-2^{1-1.0001}}{1.0001-1}=0.693123$$ $$\frac{1-2^{1-1.00001}}{1.00001-1}=0.693145$$ $$\frac{1-2^{1-1.000001}}{1.000001-1}=0.693147$$ $$\frac{1-2^{1-1.0000001}}{1.0000001-1}=0.693147$$ $$\log (2)=0.693147$$ $\endgroup$ – Mats Granvik Oct 18 '13 at 12:34
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    $\begingroup$ No, I can't explain anything. Edit: There is a way of matrix multiplication (or row sums of a lower triangular array) which explains it in terms of the Riemann zeta function at the pole, which connects it to known results. math.stackexchange.com/questions/46378/… $\endgroup$ – Mats Granvik Oct 18 '13 at 12:50
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    $\begingroup$ @bluesh34: If $f(x)=2^x$, Mats Granvik's quotients are successive approximations to $f'(0)=\ln(2)$, taking $h=-0.1$, $h=-0.01$, etc. $\endgroup$ – Andrew D. Hwang Oct 18 '13 at 13:45
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    $\begingroup$ @MatsGranvik The problem is to compute $2^{-0.01}$; how'd you do it with a pocket calculator (or by hand)? $\endgroup$ – egreg Oct 18 '13 at 14:00

10 Answers 10


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

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Image source


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

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    $\begingroup$ The series for $\log(2)$ converges very slowly. It is out of question to compute it by hand. The other formula evaluated at $x=\frac13$ is a little better. $\endgroup$ – Yves Daoust Jan 7 '16 at 10:34

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 $0.7$.


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.

  • 1
    $\begingroup$ It's also worthwhile to note that for the stated series (modulo the missing factor of $(-1)^{k-1}$), the convergence is much faster by taking $x = 1/2$ (and using $\ln(1/2) = -\ln(2)$) than by setting $x = 2$. :) $\endgroup$ – Andrew D. Hwang Oct 18 '13 at 13:29


The denominator is $(2k+1)9^k$.


Gourdon and Sebah discuss the efficiency of this formula in http://plouffe.fr/simon/articles/log2.pdf (page 11)

A "little more effort" is required to compute $log(2)$ using this formula than to compute $\pi$ using Machin's relation.

  • $\begingroup$ See the more general answer by Boris Novikov. $\endgroup$ – Yves Daoust Jan 7 '16 at 10:35
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    $\begingroup$ Setting $x=\frac{1}{\sqrt b}$ into the expansion of $atanh(x)$ yields $$\frac{\sqrt b}{2}log\left(\frac{\sqrt b+1}{\sqrt b−1}\right)=\sum_{k=0}^\infty \frac{1}{(2k+1)b^k}$$ (oeis.org/A171220 and formula 100 in crd-legacy.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf) $\endgroup$ – Jaume Oliver Lafont Jan 8 '16 at 5:35

We have the CORDIC method, which can be quite effective for by-hand computation as it requires additions/subtractions only (an one multiply by a small integer).

There are two limitations though:

  • it is better performed in base $2$, so a preliminary change of base is needed for the input argument (you can do it in base $10$ as well but it takes about $3$ times more operations);

  • you need a small table of constants.

It is based on the identity $\log(ab)=\log(a)+\log(b)$.

You first normalize the binary number as $x=z\cdot2^e$, with $1\le z<10_b$. You have $\log(x)=\log(z)+e\cdot\log(2)$.

Then $$\log(z)=\log(0.11_bz)-\log(0.11_b)\\ \log(z)=\log(0.111_bz)-\log(0.111_b)\\ \log(z)=\log(0.1111_bz)-\log(0.1111_b)\\ \cdots$$

You will use these equalities as follows. Initialize an accumulator $l\leftarrow0$ and

if $0.11_bz>1$ (i.e. $z>1.01010101_b\cdots$) let $z\leftarrow 0.11_bz$, $l\leftarrow l-\log(0.11_b)$;

if $0.111_bz>1$ (i.e. $z>1.00100100_b\cdots$) let $z\leftarrow 0.111_bz$, $l\leftarrow l-\log(0.111_b)$;

if $0.1111_bz>1$ (i.e. $z>1.00010001_b\cdots$) let $z\leftarrow 0.1111_bz$, $l\leftarrow l-\log(0.1111_b)$;


The multiplies are actually performed as shifts and subtractions (f.i. $0.111_bz=z-0.001_bz$).

This way, we progressively reduce $z$ to bring it closer and closer to $1$, while $l$ gets closer and closer to the logarithm of the initial $z$. On every step we gain one bit of accuracy.

The table of constants ($\log(10_b)=-\log(0.1_b),-\log(0.11_b),-\log(0.111_b),\cdots$ up to the desired number of significant bits) is computed in the decimal base, so that the answer is readily available as such.

$$\begin{align}z&\to-\log(z)\\ 0.1000000000000000000000000000000_b&\to 0.6931471806_d\\ 0.1100000000000000000000000000000_b&\to 0.2876820725_d\\ 0.1110000000000000000000000000000_b&\to 0.1335313926_d\\ 0.1111000000000000000000000000000_b&\to 0.0645385211_d\\ 0.1111100000000000000000000000000_b&\to 0.0317486983_d\\ 0.1111110000000000000000000000000_b&\to 0.0157483570_d\\ 0.1111111000000000000000000000000_b&\to 0.0078431775_d\\ 0.1111111100000000000000000000000_b&\to 0.0039138993_d\\ 0.1111111110000000000000000000000_b&\to 0.0019550348_d\\ 0.1111111111000000000000000000000_b&\to 0.0009770396_d\\ 0.1111111111100000000000000000000_b&\to 0.0004884005_d\\ 0.1111111111110000000000000000000_b&\to 0.0002441704_d\\ 0.1111111111111000000000000000000_b&\to 0.0001220778_d\\ 0.1111111111111100000000000000000_b&\to 0.0000610370_d\\ 0.1111111111111110000000000000000_b&\to 0.0000305180_d\\ 0.1111111111111111000000000000000_b&\to 0.0000152589_d\\ 0.1111111111111111100000000000000_b&\to 0.0000076294_d\\ 0.1111111111111111110000000000000_b&\to 0.0000038147_d\\ 0.1111111111111111111000000000000_b&\to 0.0000019074_d\\ 0.1111111111111111111100000000000_b&\to 0.0000009537_d\\ 0.1111111111111111111110000000000_b&\to 0.0000004768_d\\ 0.1111111111111111111111000000000_b&\to 0.0000002384_d\\ 0.1111111111111111111111100000000_b&\to 0.0000001192_d\\ 0.1111111111111111111111110000000_b&\to 0.0000000596_d\\ 0.1111111111111111111111111000000_b&\to 0.0000000298_d\\ 0.1111111111111111111111111100000_b&\to 0.0000000149_d\\ 0.1111111111111111111111111110000_b&\to 0.0000000075_d\\ 0.1111111111111111111111111111000_b&\to 0.0000000037_d\\ 0.1111111111111111111111111111100_b&\to 0.0000000019_d\\ 0.1111111111111111111111111111110_b&\to 0.0000000009_d\\ 0.1111111111111111111111111111111_b&\to 0.0000000005_d\\ \end{align}$$

  • $\begingroup$ This is one of the rare approaches that is workable by hand to get significant accuracy. Other methods based on series involve the computation of powers of the argument, which is a pain in the neck. Before calculators, multiplication of decimal numbers by hand was so tedious that they were accelerated through the use of... tables of logarithms (an before these were invented, trigonometric tables and the prosthaphaeresis method). $\endgroup$ – Yves Daoust Jan 8 '16 at 10:07

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. ;-))

  • $\begingroup$ The task was to avoid a pocket calculator $\endgroup$ – Hagen von Eitzen Oct 18 '13 at 15:30
  • $\begingroup$ @HagenvonEitzen One can compute square roots by hand. ;-) For this you need a pocket calculator with the four operations and the square root, not a full fledged calculator. With only four square roots ($n=16$) I get $0.67834750882272$; with eight ($n=256$) it's $0.69220964211968$. This is easier than using the Taylor expansion. Completely by hand: Taylor in the form $\log\frac{1+x}{1-x}$ $\endgroup$ – egreg Oct 18 '13 at 15:43

$$\log (x)=\sum _{n=1}^{\infty } \frac{\left(\frac{x-1}{x}\right)^n}{n}$$ when $x>1$


What you can use is the Taylor expansion of $\ln (1+x)$:

$$\ln (1+x) = \sum (-1)^{j+1}{x^j\over j}$$

which converges for $-1<x\le1$. It would be tempting to insert $x=1$ into it, but that would be a poor choice since the convergence for $x=1$ is painfully slow. Instead you use the fact that $\ln 2 = -\ln 1/2$ and insert $x=-1/2$ instead:

$$\ln (1-{1\over 2}) = \sum (-1)^{j+1}{1\over j2^j} = -\sum {1\over j2^j}$$


$$\ln 2 = \sum {1\over j2^j}$$

This is similar to how the calculator does it, but there's probably a few tricks more that's used. First it probably uses base two logarithm and have a stored value of $\lg_2 e$ to be able to produce the natural logarithm. The reason for this is to be able to handle logarithm of values outside the convergence region (and generally we want to use the series for as narrow region as possible). We generally can write any number on the form $x2^p$ (in fact the numbers are already represented on that form) with $x$ being near $1$ and then $\lg_2(x2^p) = p\lg_2(x)$ (similar trick is being done on all these kind of functions).

The second trick is to approximate $\ln(1+x)$ on the interval $[1/\sqrt2, \sqrt2]$ even better than Taylor expansion, the trick is to find a polynomial that approximates it as uniformly good as possible. The McLaurin expansion has the property that it will yield a good approximation fast for values near zero at the expense of values further away. For generic case one uses a polynomial that yields a good enough approximation equally fast in the interval.


We can represent the logarithm of positive rational numbers as follows.

First, consider the following null conditionally convergent series (cancelled harmonic series):


Note that we are computing $0=log(1)=log\left(\frac{1}{1}\right)$ by adding consecutive terms with 1 positive fraction and 1 negative fraction each, taken from the inverses of non-zero integers. This observation may sound trivial now, but it is interesting for what comes next.

We can rearrange the terms of this series to compute $log(2)$ by taking two positive fractions and one negative for each term.


This can be easily seen to be the Mercator series in disguise, so we have discovered nothing new yet.

But there is more. Similarly, we have


This pattern holds for all positive integers, so the next step is applying the property that $log(p/q)=log(p)-log(q)$ on these representations.

This leads to $log(p/q)$ by adding $p$ positive fractions and $q$ negative fractions at each step. For example, we have


as illustrated in http://oeis.org/A166871.

See also Do these series converge to logarithms? and Series for logarithms for further discussion of generalized Mercator series.

  • $\begingroup$ Would you really advise this for by-hand computation ? $\endgroup$ – Yves Daoust Jan 7 '16 at 10:36
  • $\begingroup$ This cannot be expected to be any faster than the better known Leibniz series for $\frac{\pi}{4}$. However, proper truncation for the logarithm of odd integers shows some improvement, so for the logarithm of the ratio of two large close odd integers the approximation may be even useful (see math.stackexchange.com/a/1602945/134791) $\endgroup$ – Jaume Oliver Lafont Jan 7 '16 at 11:09

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