I have read through questions like Calculate logarithms by hand and and a section of the Feynman Lecture series which talks about calculation of logarithms. I have recognized neither of them useful for my purpose which is to quickly calculate logarithms of base $10$ upto $4$ digit accuracy
(I believe 4 is the goldilocks number in this case) .

I wish to find things like $\log_{10}(2) \approx 0.3010$ quickly without using a calculator or log table. Why? Because I want to be free from carrying them around and losing them all day. Plus, they're not always available when I need them (you can guess why). My main purpose is to approximate the answers of very large and very small results of time consuming calculations. Logarithms make that job much easier for me. For example,

$$\frac{87539319}{1729} \approx 10^{7.942 - 3.237} = 10^{4.705} = 5.069*10^4$$

According to Wolfram (Yup, I'm that lazy) the answer is, $50630.0\overline{283400809716599190}$. Yes, I've over estimated by around $60$ but thanks to a log table, I did that approximation as fast as it took Wolfram to load the precise answer in my browser. But, without a log table, dividing itself would have me executing an iterative convergence just to find the multiples.
(1729*2 = too low, 1729*8= too high ... this must be so intuitive for most of you)

So, a quick approximation method for logarithms would be really helpful to me.

Also, a good way to find antilogs will be nice as well. I just realized that I can't compute decimal powers. $$\Large 10^{0.3010} = 10^{0.3}*10^{0.001} = \sqrt[10]{1000} * \sqrt[1000]{10} = \text{Calculator Required}$$ I checked " How to calculate a decimal power of a number" but alas, the thing which came closest to what I needed required a calculator for an intermediate step. Defeats the purpose, I know.
If I can't find the antilog, the whole point of having a quick way to find the logarithm would be lost.

I hope you can help.

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    $\begingroup$ They're a little before my time, but I think some of the larger/more detailed slide rules could do something like this. $\endgroup$ Mar 4, 2014 at 4:36
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    $\begingroup$ Also, you might find something interesting in this thread. $\endgroup$ Mar 4, 2014 at 4:51

6 Answers 6


For quick-and-dirty approximations, I'm fond of the musical logarithms (writeup by Sanjoy Mahajan of work due to I. J. Good, who credits his father). Essentially this boils down to a mathematical fact:

  • $2^{10} \approx 10^3$, and taking 120th roots, $2^{1/12} \approx 10^{1/40}$.

and a "musical" fact:

  • many rational numbers with small numerator and denominator can be approximated as powers of $2^{1/12}$.

I call this a "musical" fact because $2^{1/12}$ is the frequency ratio corresponding to an (equal-tempered) semitone.

For example: $3/2$ is the frequency ratio corresponding to the musical interval of a perfect fifth, which is seven semitones; thus $3/2 \approx 2^{7/12} \approx 10^{7/40}$ and so $\log_{10} 3/2 \approx 7/40$.

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    $\begingroup$ This is so beautiful :D $\endgroup$
    – Nick
    Sep 6, 2014 at 6:06

As you want to do calculations by hand, you are merely asking how to build a table of logarithms.

There are at least two ways:

  • Computing $\sqrt{10}$, then the square root of it, etc. until the result is small enough to do an approximation. That's described in Feynman lecture.
  • Using series, for example $\ln (1+t)=t-\frac12t^2+\frac13t^3+\cdots$. There is a way to use them to compute approximations of $\ln k$ for $k\in\{2,3,5,7\}$. With them you compute $\ln 10$ and its inverse, and it helps you compute decimal logs from natural log. Then by applying the series to fractions $\frac{n+1}{n}$, you can, step by step, compute all logarithms you need.

I can give some more details if necessary.

Combining the series for $\ln(1+t)$ and $\ln(1-t)$, you get a series with only odd terms, so it's faster to use.

There are also methods to compute logarithms of trigonometric functions: trigonometric formulas are very useful to simplify many computations by logarithms.

There are also special tables, like a $20$ decimals one in Hoüel tables of logarithms, to compute the log or antilog of a number to $20$ decimal places, with only a very small table (one page) with logs of numbers $1+d\cdot10^k$.

Now, all of these need an unbelievable amount of time to build a table of logarithms, even if you want only logs of numbers from $100$ to $1000$ to four decimals. And there is no simple way to compute just one logarithm only by hand. There are the series above, polynomial approximations, and probably other ways, but nothing as easy to use as a table. And algorithms used by computers are far from usable by hand.

So, if you insist on not using a calculator or a computer, which I perfectly understand, I still wonder why you want to do what took decades to mathematicians of the past to build reliable tables. Doing it yourself would be an intersting endeavor, but I can't imagine it's only to have the table at hand: then, why not simply use a well known table, like those by Schrön, Callet or Dupuis, among others? Doing it yourself, and alone, is the best way to make many computational mistakes a get an unreliable table. Notice that there were mistakes even in well-known tables (I found a list of errata for Schrön's table for example), and the verification step is certainly not the least important.

Notice old tables are still available as used books (the latest were published in the eighties, and are in very good condition, but even old ones from the late 19th century are in fairly good shape). You can even find slide rules - even new ones, as I bought several on Faber-Castell's site, never open (very old stock I guess). There are also many scanned tables on the net, and this nice site.

I'd be glad to develop on any part, but I would like to understand what you really want. And notice, if there was a usable trick by hand with no table, no slide rule, and nothing but a pen and paper, it would have been used instead of the heavyweight tables :-) It's not by luck or magic they were so widespread before the appearance of calculators.

(too long for a comment)

@Nick There is something else I didn't read at first: "free myself from log tables and calculators which are not at my disposal during exams". An exam without any computing tool that would ask computations would be... strange. But not unbelievable, as I saw more or less this in France at the chemistry test in competitive examinations at the end of classes préparatoires. IIRC, only a tiny (1/4 of a sheet) table of logarithms was given to compute numerical results. Believe it or not, some schools still allow slide rules for these exams (and only a couple of years ago, also tables of logarithms, I mean, the "real" ones, not the one sheet ersatz). Two examples I know of are the Mines-Ponts exam, and the École Polytechnique: it's still explicitly stated in the exam rules as of 2014 exams. However, virtually no student is coming to the exam with a slide rule, and I bet only few would know what it is, let alone how to use it appropriately.

  • $\begingroup$ Actually, in Physics, the subject which requires logs for calculating, most errors are alright. It's disappointing to hear : "...there is no simple way to compute just one logarithm only by hand..." . This was what I wanted. But that's alright, I appreciate your patience and understanding. I just needed one concrete way to approximate logs which gets a close answer most of the time so as to roughly attempt certain calculations. I mean, with intuition and common sense, how big of an error can I really make? Tables won't be around when I'm shipwrecked or something. Imagine that :D $\endgroup$
    – Nick
    Aug 31, 2014 at 18:42
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    $\begingroup$ @Nick The best you can do, I think (I mean the most workable by hand), is giving a polynomial approximation of the decimal log on $[1,10]$, and using the fact that $\log a\cdot10^n=n+\log a$. It won't be extra accurate, but if it's enough for you, then why not. $\endgroup$ Aug 31, 2014 at 18:48
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    $\begingroup$ @Nick As a follow-up about polynomial approximation, if you have Abramowitz & Stegun at hand, go have a look at page 68. If you haven't it, it's free and here. There is a large litterature about approximations of functions, and espacially polynomials. And polynomials are probably the most manageable by hand. $\endgroup$ Aug 31, 2014 at 18:53
  • $\begingroup$ @Nick One more comment, even if it repeats a bit the answer: the preceding polynomial approximation is still more difficult to useby hand than a good old table of logarithms, which usually requires only a linear interpolation, and even the linear interpolation is rendered extremely easy by the small proportion tables in most tables. And there are very small tables of logarithms (~ A6 format), for 5 places (1000-10000), so they really don't take too much space. And for 4 places (100-1000), one sheet is enough. And there are also small calculators, so... $\endgroup$ Aug 31, 2014 at 19:06
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    $\begingroup$ @Nick I didn't see your last sentence. shipwrecked? I guess logs would be the least of your problems :-) But in the same vein, I believe navigators still must get a table of logs with them, together with mechanical instruments, in case GPS electronics goes down. There are the Norie's tables for example. (pdf of older edition (1836) here) $\endgroup$ Aug 31, 2014 at 19:17

To give an approximation for at least $4$ digits in general by hand I think it is almost impossible. If you know some results from approximation theory after that you can appreciate logarithm tables.

Of course the first idea is the Taylor expansion for few terms. We know that for $|x| \leq 1$ and $x \neq -1$ the series for $\ln(1+x)$ is the following. $$ \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots $$ You can "run it" by hand from $n=1 \dots 3$ and for $\ln(2)$ you get $0.8333333333$. The correct value for $\ln(2)$ is $0.6931471806$. So the problem is behind the rate of convergence. There are also importent domain restrictions for this method.

For small $x$ values we also know that $\log(x) \approx \frac{x^x-1}{x}$ and $\log(1+x) \approx x$. Which is also not a good approximation but we can use it for values less then $1$. With logarithm identitiy-tricks you can make it more accurate, but we have better solutions.

Now take a look at inequalities. We have that for all $x>0$: $$1-\frac{1}{x} \leq \ln x \leq x-1.$$ Or we can write it to the form for all $x>-1$: $$\frac{x}{1+x} \leq \ln(1+x) \leq x.$$

We know other inequalities, and I think it is a good approach by hand, so let me introduce Henri Padé and his Padé approximant. With this method you can give lower and upper bounds for a function with rational functions. We will call lower bound $\phi_n$ and upper $\psi_n$, and here $n$ is the order of the approximation. You can read about approximate $\ln(1+x)$ with this method in this really good paper, or in this website. So we have $$\phi_n(x) \leq \ln(1+x) \leq \psi_n(x)$$ for $x \in [0,\infty[$ and for each $n$.

We will take order $n=3$, because I think this two rational funcion is what we can handle by hand. If you are good at mental calculation and you can memorize functions easily, you can take higher orders from the paper I refered above. So for the lower bound $\phi_3$ we get $$\phi_3(x)=\frac{x(60+60x+11x^2)}{3(20+30x+12x^2+x^3)},$$ and for the upper bound $\psi_3$ we get $$\psi_3(x)=\frac{x(30+21x+x^2)}{3(10+12x+3x^2)}.$$

To evaulate this two functions by hand you need just to add, multiply, divide, and take integer power of a number.

To see how accuare this method I give you some results.

  • $\phi_3(1) = 0.6931216931 \leq \ln(2) = 0.6931471806 \leq 0.6933333333 = \psi_3(1),$
  • $\phi_3(2) = 1.098039216 \leq \ln(3) = 1.098612289 \leq 1.101449275 = \psi_3(2),$
  • $\phi_3(3) = 1.383673469 \leq \ln(4) = 1.386294361 \leq 1.397260274 = \psi_3(3),$
  • $\phi_3(4) = 1.602693603 \leq \ln(5) = 1.609437912 \leq 1.635220126 = \psi_3(4),$
  • $\phi_3(9) = 2.246609744 \leq \ln(10) = 2.302585093 \leq 2.493074792 = \psi_3(9),$
  • $\phi_3(50) = 3.254110231 \leq \ln(51) = 3.931825633 \leq 7.357172215 = \psi_3(50).$

Of course because the method works for smaller $x$ values better, if you have large $x$, then you could combine Padé approximant with logarithmic identities. For example $51$ has the prime factors $3$ and $17$, because of that we can write $\ln(51)$ into the form $\ln(51)=\ln(3)+\ln(17)$ so $$\phi_3(50) \leq \phi_3(2)+\phi_3(16) = 3.766096945 \leq \ln(51)$$ is a better lower bound, and $$\ln(51) \leq \psi_3(2)+\psi_3(16) = 4.521380547 \leq \psi_3(50)$$ is a better upper bound.

This is also a good approach to get approximation for $\log_b(x)$. For example for $\log_{10}(2) = \ln(2) / \ln(10) = 0.3010299957$ we can say it is somewhere between $\psi_3(1) / \psi_3(9) = 0.2781037037$ and $\phi_3(1) / \phi_3(9) = 0.3085189562$.

And at last if you get an $n=5$ order Padé approximant and use logarithmic identities then you get the following approximation for $\log_{10}(2)$ with $\phi_5$.

$$\frac{\phi_5(1)}{\phi_5(1)+\phi_5(4)} = 0.3010494871,$$

which is correct for the first $4$ digits.

You can approximate exponential function with this method too. Read about it in this paper, or in this MathOverflow answer!

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    $\begingroup$ I like this answer because it's realistic and I see myself applying it. You deserve the best answer (although the bounty will go to Jean-Claud's informative rant). Anyhow, thank you for introducing me to this method. Although I won't be using it in any exams, it will surely be of use to me someday. Again, thank you. :D $\endgroup$
    – Nick
    Sep 6, 2014 at 6:22

(Disclaimer: I'm new to stack exchange, so I apologize for any breaches of etiquette. I read that bumping old posts is permitted here, but perhaps people disagree with that or it's since been changed? I thought my method might be useful to others because it's the simplest one I've seen and I haven't seen it anywhere else)

Onto the math: Calculating logarithms by hand has been an interest of mine lately, and my method is quite simple. Accuracy is typically about 4 digits. I don't think I can explain it better than I can show it, so I'll use your example of log 1729.

First, we break it down. log 1729= log 1000 + log 1.729 log 1000=3 so log 1729= 3 + log 1.729

log 1.729= ln 1.729/ln 10

ln 1.729= ln 1.65 + ln (1.729/1.65)

ln 1.65 is approximately 0.5, so ln 1.729= 0.5 + ln (1.729/1.65)

(Yes, this does require you to be familiar with some of the natural logarithms, such as ln 1.65 being about 0.5-- see the end for more about this)

ln 10= ln (7.39) + ln (10/7.39) ln 10= 2 + ln (10/7.39) (approximately)

If we go back and plug things in, we get that log 1.729 approximately equals:


Here's the trick: When x is between 1 and 1.65, ln x is roughly equal to (x^2 -1)/(2x)

Now, by plugging in x= 1.729/1.65, we get roughly 0.046784964, thus making ln 1.729 roughly equal to 0.546784964 (if you're concerned about the accuracy, the error margin will get smaller when we divide the two natural logs)

Doing this same process, ln(10/7.39) is roughly 0.307089986, making ln 10 roughly 2.307089986

Then, if we divide them, we get our approximation of log 1.729: 0.237002010

Thus, log 1729 is roughly 3 + 0.237002010, or about 3.237

In summary, the method is to:

  • Split the original log (like scientific notation)
  • Use the change of base formula on the remainder
  • Split the numerator/denominator of the remainder until the only part inside the logarithm is between 1 and 1.65
  • Apply the formula to the part still in the logarithm
  • Simplify

It would be dishonest not to mention the areas where this process is lacking. Here are the major caveats:

  • Requires memorization of natural logs: it's really not very hard to do, and is much easier than memorizing the log tables, but the point remains. If you're using base 10 logarithms exclusively, then there's really not very many numbers to memorize. Using this method, when you apply the change of base formula you shouldn't have to find the natural log of any number greater than 10. That means you really only need to memorize four: ln(1.65), ln(2.72), ln(4.48), and ln(7.39). Of course, you could pick different values but I think those are the easiest.
  • Accuracy: Although you can expect this method to typically be accurate to 4 digits, it wouldn't surprise me if it's off by more at times. To increase accuracy, basically just use longer decimals (you may want to extend the natural logs I mentioned above to three digits). Also, the formula works best for smaller values-- x=1.2 will be more accurate than x=1.6.
  • Difficulty of dividing decimals: sure, you could divide those decimals by hand, but who would want to? This process might be more difficult than just dividing out the numbers in your example. This process is probably most useful for a situation where you have a calculator that doesn't have a log button. I think it's unlikely that you'll find a much easier and general way of calculating logs.

Well, this got long quickly. Hopefully this can help you or anybody else who happens to stumble upon the thread like I did.

  • $\begingroup$ Alternatively, if you know that $\log 2 \approx 0.301$ and $\log 3 \approx 0.477$, then we have $\log 1729 \approx \log 1728 = 3 \log 12 = 3 (2 \log 2 + \log 3) \approx 3.327 $. $\endgroup$
    – Théophile
    Feb 6, 2023 at 14:48

Jacques Laporte has a page explaining some algorithms that work digit by digit. For other functions (e.g. trigonometric and hyperbolic) there is the class of CORDIC algorithms. Such algorithms were used in the first HP calculators, as they need very modest hardware.

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    $\begingroup$ As a highschooler, it's difficult to interpret the answer I need from Jacques page. $\endgroup$
    – Nick
    Aug 31, 2014 at 15:22

Although this is not exactly what you ask for, if you have a calculator with square root, you can use the same method Briggs used in 1620 (before calculus, integration, and power series). Take the square root of your number repeatedly until you get a 1 followed by some zeros and a set of digits. Stop when the number of zeros and the number of digits are about the same. Subtract the 1 and multiply by the appropriate power of 2 (the number of times you hit square root), and you are done. Keep in mind that for Briggs, extraction of square root was a tedious affair; for you, it is a single finger poke.

The bad news is that you will have to memorize a constant to convert this natural log to base 10. But you only have to memorize half of it because it is 0.4343. I used this method in 1973 when an engineer at work got a box of defective five banger calculators. He fixed them by cleaning the key contacts and sold them for $3. I needed the more precise logs than a slide rule provided for decibel calculations.

There are refinements that have been touched on by others here, such as 2(x-1)/(x+1), but that doesn't address your problem of working by hand.

I recommend that you memorize the above constant, Log(e), and also the logs of 2 and 3. I assume that you have already memorized the log of 10 to one billion decimal places. From these factors, you can construct a number that is close to your target. But this gets to be more work than the simple division problem you need to address.

For crude work, I memorized the numbers whose logarithms are 0.1, 0.2, etc., which are the power ratios for 1, 2, etc. decibels. Eyeball interpolation gets me the second digit of the log. Not as good as a slide rule or the above square root method, but better than nothing when many calculations must be chained. I am presently building a slide rule with a spiral scale 50 turns around a 2" pipe, inspired by the Otis King calculator. It is easy with a Microsoft excel chart. I think it would give you the four digits you seek, but it is not a portable device.


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