Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I replace the $\log(x)$ function by simple math operators like $+,-,\div$, and $\times$?

I am writing a computer code and I must use $\log(x)$ in it. However, the technology I am using does not provide a facility to calculate the logarithm. Therefore, I need to implement my own function using only simple operators ($+,-,\div$, and $\times$).

Thank you.

share|cite|improve this question
How about Taylor series? – user12205 Oct 23 '11 at 12:01
The best you can do is approximating the logarithm using its power series expansion. – Rasmus Oct 23 '11 at 12:02
You could probably use this. – user12205 Oct 23 '11 at 12:02
There is no way to write $\log x$ as (finitely many) simple math operations applied to $x$. You will need to approximate the logarithm using any of several methods; lists some of these. – user7530 Oct 23 '11 at 12:05
Any $x>0$ is of the form $x=10^r\cdot t$ with $r\in{\mathbb Z}$ and $1\leq t<10$. I suggest that you use a ("wired in") value for $\ln 10$ and a good rational approximation for $t\mapsto\ln t$ in the interval $1\leq t\leq 10$. – Christian Blatter Oct 23 '11 at 13:31

Contrary to popular belief, you can do better that power series. The trick is in the use of continued fractions and the related Padé approximants.

One continued fraction for the logarithm (due to Khovanskiĭ) goes a bit like this:


The beauty of this is that it has a wider domain of applicability: it is valid as long as $|\arg(1+z)| < \pi$.

One can use the Lentz-Thompson-Barnett method on this CF, of course, but one could also choose to exploit argument reduction here, by suitably exploiting the identity $\log(ab)=\log\,a+\log\,b$. If you take that route, you can be justified in just using a truncation of the continued fraction. That truncation is what's called a Padé approximant.

I'll edit later with more details if needed.

share|cite|improve this answer

In the interest of demonstrating that there is more than one way to skin a cat, I display here Borchardt's algorithm for computing $\log\,x$, which is a modification of the more conventional arithmetic-geometric mean iteration:

$a_0=\dfrac{1+x}{2};\quad b_0=\sqrt{x};$


$\text{until }|a_k-b_k| < \varepsilon$
$\log\,x\approx 2\dfrac{x-1}{a_k+b_k}$

This of course presumes that you have a square root function available, but that remains doable even with your restrictions.

If you find that the convergence rate is too slow for your taste, Carlson shows that you can use Richardson extrapolation to speed up the convergence in the article I linked to. From my own experiments, I've found that the convergence rate of the unmodified Borchardt algorithm is already pretty decent, so unless you do need the speed-up (and remembering that Richardson extrapolation requires an auxiliary array for its implementation, which adds to the storage cost), vanilla Borchardt is fine as it is.

share|cite|improve this answer
+1 for the algorithm! – user12205 Oct 24 '11 at 23:43

Curiously, nobody proposed the CORDIC algorithm that was very useful when the 'price' of multiplication and division was high and/or the CPU limited.
The trick is to use a precomputed table of logarithms (say $\ln(10), \ln(2), \ln(1.1), \ln(1.01)... \ln(1.000001)$) and compute any logarithm using only addition/subtraction and shift operations (code example here).
A little late but...

share|cite|improve this answer
As always, there is always the trade-off between speed and storage... – J. M. Dec 30 '11 at 14:15
@Guesswhoitis: CORDIC takes one floating-point constant per bit, not really a big deal. The corresponding storage could easily be exceeded by the code of more complicated algorithms. – Yves Daoust Jun 7 '15 at 13:26

If number $N$ (base 10) is $n$-digit then

$$n-1 \leq \log_{10}(N) < n$$

Then logarithm can be approximated using

$$\log_{10}(N) \approx n-1 + \frac{N}{10^{n} - 10^{n-1}}$$


$\log_{10}(53) = 1.72427587 $

Here $n= 2, N=53$ then,

$$\log_{10}(53) = 2 -1 + \frac{53}{100-10}=1.588888$$

Logarithm maps numbers from 10 to 100 in the range 1 to 2 so log of numbers near 50 is about 1.5. But this is only a linear approximation, good for mental calculation and toy projects but not that good for serious research.

share|cite|improve this answer

The Wikipedia article Generalized continued fraction has a Khovanskiĭ-based algorithm that differs only in substituting $x/y$ for $z$, and showing an intermediate step:

$$ \log \left( 1+\frac{x}{y} \right) = \cfrac{x} {y+\cfrac{1x} {2+\cfrac{1x} {3y+\cfrac{2x} {2+\cfrac{2x} {5y+\cfrac{3x} {2+\ddots}}}}}} = \cfrac{2x} {2y+x-\cfrac{(1x)^2} {3(2y+x)-\cfrac{(2x)^2} {5(2y+x)-\cfrac{(3x)^2} {7(2y+x)-\ddots}}}} $$

share|cite|improve this answer

Here I use log to mean logarithm base 10.

Here is a quick, iterative method to compute $\log x$ for any $1 \le x \le 10.$

[INITIALIZE] Let $n = 0$. Define

$$\begin{array}{ccc} xl_0 = 1& xm_0=\sqrt{10} & xr_0=10 \\ yl_0 = 0& ym_0=0.5 & yr_0=1 \end{array}$$

[ITERATE] Compare $xm_n$ to $x$. If they satisfy your favorite criterion for "close enough", then $\log x = ym_n$ and we are done. Otherwise compute the following and then assign $n\to n+1$.

If $xm_n < x$,

$$\begin{array}{ccc} xl_{n+1} = xl_n& xm_{n+1}=\sqrt{xl_n \cdot xm_n} & xr_{n+1}=xm_n \\ yl_{n+1}=yl_n& ym_{n+1}=(yl_n+ym_n)/2 & yr_{n+1}=ym_n \end{array}$$

If $xm_n > x$,

$$\begin{array}{ccc} xl_{n+1} = xm_n& xm_{n+1}=\sqrt{xm_n \cdot xr_n} & xr_{n+1}=xr_n \\ yl_{n+1}=ym_n& ym_{n+1}=(ym_n+yr_n)/2 & yr_{n+1}=yr_n \end{array}$$

This is an extremely simple program to write and it returns reasonably accurate values of $\log x $ for $1 \le x < 10$. If you need $\ln x$, just use $\ln x = \dfrac{\log x}{\log e}$

You might also find THIS interesting. Just scroll down to "An Algorithm For Logarithms".

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.