Approximating Logs and Antilogs by hand

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!

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.

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@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.

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 $\log_{10} 3/2 \approx 7/40$.