Mentally compute logs base 2

Posted on 29 October 2025 by John

The previous post required computing

$\frac{128}{\log_2 5}$

After writing the post, I thought about how you would mentally approximate log₂ 5. The most crude approximation would round 5 down to 4 and use log₂ 4 = 2 to approximate log₂ 5. That would be good enough for an order of magnitude guess, but we can do much better without too much more work.

Simple approximation

I’ve written before about the approximation

$\log_2 x \approx 3\frac{x - 1}{x + 1}$

for x between 1/√2 and √2. We can write 5 as 4 (5/4) and so

$\begin{align*} \log_2 5 &= \log_2 4 (5/4) \\ &= \log_2 4 + \log_2 (5/4) \\ &\approx 2 + 3\frac{5/4 - 1}{5/4 + 1} \\ &= 2 + 3 \frac{1/4}{9/4} \\ &= 7/3 \end{align*}$

How accurate is this? The exact value of log₂ 5 is 2.3219…. Approximating this number by 7/3 is much better than approximating it by just 2, reducing the relative error from 16% down to 0.5%.

Origin story

Where did the approximation

$\log_2 x \approx 3\frac{x - 1}{x + 1}$

come from?

I don’t remember where I found it. I wouldn’t be surprised if it was from something Ron Doerfler wrote. But how might someone have derived it?

You’d like an approximation that works on the interval from 1/√2 to √2 because you can always multiply or divide by a power of 2 to reduce the problem to this interval. Rational approximations are the usual way to approximate functions over an interval [1], and for mental calculation you’d want to use the lowest order possible, i.e. degree 1 in the numerator and denominator.

Here’s how we could ask Mathematica to find a rational approximation for us [2].

Simplify[
    N[
        ResourceFunction["EconomizedRationalApproximation"][
            Log[2, x], { x, {1/Sqrt[2], Sqrt[2]}, 1, 1}]]]

This returns

(2.97035 x − 2.97155) / (1.04593 + x)

which we round off to

(3 x − 3) / (1 + x).

The N function turns a symbolic result into one with floating point numbers. Without this call we get a complicated expression involving square roots and logs of rational numbers.

The Simplify function returns an algebraically equivalent but simpler expression for its argument. In our case the function finishes the calculation by removing some parentheses.

[1] Power series approximations are easier to compute, but power series approximations don’t give the best accuracy over an interval. Power series are excellent at the point where they’re centered, but degrade as you move away from the center. Rational approximations spread the error more uniformly.

[2] I first tried using Mathematica’s MiniMaxApproximation function, but it ran into numerical problems, so I switched to EconomizedRationalApproximation.

Day of the week centuries from now

Posted on 18 September 2025 by John

Yesterday I wrote a post outlining mental math posts I’d written over the years. I don’t write about mental math that often, but I’ve been writing for a long time, and so the posts add up. In the process of writing the outline post I noticed a small gap.

In the post on mentally calculating the day of the week I focus on this century, then tersely say “For dates in the 20th century, add 1. For dates in the 22nd century, subtract 2.” This post will expand on that. I’ll say a little more about what these rules mean, how to extend them to other centuries, and why they work.

Dates in the 20th century

Suppose you wanted to know the day of the week for July 4, 1976. You could use the method in the post linked above and end up with Saturday. Moving ahead one day of the week gives you Sunday.

Dates in future centuries

For days in the 2100s you would find the day of the week for the corresponding date in this century, then subtract two days. For the 2200s you’d add three days, and for the 2300s you’d add one day.

The Gregorian calendar repeats every 400 years, so if you can find the days of the week for the years 2000 through 2399, you can find the day of the week for all years in the future by subtracting enough multiples of 400 to get into that range of years. For example, in the year 2825, September 18 will fall on a Thursday because it falls on a Thursday today in 2025.

You could also go backward by centuries as well as forward, but there’s a complication. If you run the Gregorian calendar backward far enough, you run into a time before the Gregorian calendar was adopted.

Why the rules work

So why do the century adjustment rules work? There’s a simple but incomplete explanation, and a more complicated complete explanation.

Partial explanation

The number of days in most centuries, i.e. those not divisible by 400, is

365 × 100 + 24 = 36524

which is congruent to 5 mod 7, which is congruent to −2 mod 7. Dates in the 2100s are 36524 days ahead of their counterparts in the 2000s, so we move back two days in the week. The same applies when going from the 2100s to the 22oos, and from the 2200s to the 2300s.

For example, September 18, 2125 is 36524 days from now—note that February, 2100 will not have a Leap Day—and so it will fall two days earlier in the week, on a Tuesday.

That explanation is mostly correct, but there’s a wrinkle. Not all dates in the 2100s are 36524 days ahead of their counterparts in the 2000s. There are indeed 36524 days between September 18, 2025 and September 18, 2125. But here are 36525 days between January 1, 2000 and January 1, 2100. The former was on a Saturday and the latter will be on a Friday. It would seem that our process is wrong, but it’s not.

Full explanation

Let’s go back and look at the algorithm for finding days of the week,

Take the last two digits of the year and add the number of times 4 divides that number [1].
Add a constant corresponding to the month.
Add the day of the month.
Subtract 1 in January or February of a leap year.
Take the remainder by 7.
Adjust for the century.

Suppose we wanted to find the day of the week for January 1, 2100. The first step gives us 0. The constant for January is 6, so at the end of step 2 we have 6. At the end of step 3 we have 7.

Now comes the key part: in step 4 we do not subtract 1 because 2100 is not a leap year. Step 5 gives us 0, i.e. Sunday. When we adjust for the century by moving back two days, we get Friday.

This explanation is admittedly tedious, but it reveals a subtle part of the algorithm: you have to carry out the algorithm, in particular step 4, for the original year. To find the day of the week for a year in the 2200s, for example, you carry out the algorithm as if it were valid for the 2200s, until you get to the last step.

[1] This is known as the year share. There are many ways to calculate it (mod 7) that are less obvious but easier to calculate mentally. It’s interesting that these methods generally look more complicated in writing, but they’re easier to carry out.

Mental math posts

Posted on 17 September 2025 by John

I’ve written quite a few posts on mental math over the years. I think mental math is important/interesting for a couple reasons. First, there is some utility in being able to carry out small calculations with rough accuracy without having stop and open up a calculator. Second, the constraints imposed by mental calculation make you look at a problem differently, often in interesting ways. An approximate mental calculation often requires more understanding than a precise machine calculation.

Divisibility and factoring

Day of the week

Transcendental functions

This page outlines how to calculate logarithms base 2, e, and 10, and their inverses, linking to other posts for details.

It also outlines computing trig functions, roots, and the gamma function. See the links below for more accurate ways to compute logarithms.

Incidentally, when I’ve posted about these approximations before, inevitably someone will say these are just Taylor approximations. No, they aren’t.

Logarithms

Miscellaneous

A mental random number generator

Posted on 10 September 2025 by John

Man thinking of random numbers

George Marsaglia was a big name in random number generation. I’ve referred to his work multiple times here, most recently in this article from March on randomly generating points on a sphere. He is best remembered for his DIEHARD battery of tests for RNG quality. See, for example, this post.

I recently learned about a mental RNG that Marsaglia developed. It’s not great, but it’s pretty good for something that’s easy to mentally compute. The output is probably better than if you simply tried to make up random numbers; people are notoriously bad at doing that. Hillel Wayne explores Marsaglia’s method in detail here.

Here’s a summary the generator in Python.

state = 42 # Set the seed to any two digit number

def random_digit():
    tens = lambda n: n // 10
    ones = lambda n: n % 10
    global state
    state = tens(state) + 6*ones(state)
    return ones(state)

Tricks for radix conversion by hand

Posted on 23 January 2025 by John

The simplest trick for converting from one base to another is grouping. To convert between base b and base b^k, group numbers in sets of k and convert one group at a time. To convert from binary to octal, for instance, group bits in sets of three, starting from the right end, and convert each group to octal.

11110010_two → (11)(110)(010) → 362_eight

For an example going the other direction, let’s convert 476 in base nine to base three.

476_nine → (11)(21)(20) → 112120_three

In general, conversion between bases is too tedious to do by hand, but one important case where it’s a little easier than it could be is converting between decimal and octal. In combination with the grouping trick above, this means you could, for example, convert between decimal and binary by first converting decimal to octal. Then the conversion from octal to binary is trivial.

The key to converting between decimal and octal is to exploit the fact that 10 = 8 + 2, so powers of 10 become powers of (8 + 2), or powers of 8 become powers of (10 − 2). These tricks are easier to carry out than to explain. You can find descriptions and examples in Knuth’s TAOCP, volume 2, section 4.4.

Knuth cites a note by Walter Soden from 1953 as the first description of the trick for converting octal to decimal.

The trick for moving between base 9 and base 10 (or by grouping, between base 3 and base 10) is simpler and left as an exercise by Knuth. (Problem 12 in section 4.4, with a solution given at the back of the volume.)

Mentally multiply by π

Posted on 31 August 2024 by John

This post will give three ways to multiply by π taken from [1].

Simplest approach

Here’s a very simple observation about π :

π ≈ 3 + 0.14 + 0.0014.

So if you need to multiply by π, you need to multiply by 3 and by 14. Once you’ve multiplied by 14 once, you can reuse your work.

For example, to compute 4π, you’d compute 4 × 3 = 12 and 4 × 14 = 56. Then

4π ≈ 12 + 0.56 + 0.0056 = 12.5656.

The correct value is 12.56637… and so the error is .00077.

First refinement

Now of course π = 3.14159… and so the approximation above is wrong in the fourth decimal place. But you can squeeze out a little more accuracy with the observation

π ≈ 3 + 0.14 + 0.0014 + 0.00014 = 3.14154.

Now if we redo our calculation of 4π we get

4π ≈ 12 + 0.56 + 0.0056 + 0.00056 = 12.56616.

Now our error is .00021, which is 3.6 times smaller.

Second refinement

The approximation above is based on an underestimate of π. We can improve it a bit by adding half of our last term, based on

π ≈ 3 + 0.14 + 0.0014 + 0.00014 + 0.00014/2 = 3.14157

So in our running example,

4π ≈ 12 + 0.56 + 0.0056 + 0.00056 + 00028 = 12.5656 = 12.56654.

which has an error of 0.00007, which is three times smaller than above.

[1] Trevor Lipscombe. Mental mathematics for multiples of π. The Mathematical Gazette, Vol. 97, No. 538 (March 2013), pp. 167–169

Antenna length: Another rule of 72

Posted on 27 March 2024 by John

The famous Rule of 72 says that to find out how many years it takes an investment to double in value, divide 72 by the annual percentage rate. I’ll come back to that in a little bit.

This morning I read a really good article, Fifty Things you can do with a Software Defined Radio. The article includes a rule of thumb for how long an antenna needs to be.

My rule of thumb was to divide 72 by the frequency in MHz, and take that as the length of each side of the dipole in meters [1]. That’d make the whole antenna a bit shorter than half of the wavelength.

Ideally an antenna should be as long as half a wavelength of the signal you want to receive. Light travels 3 × 10⁸ meters per second, so one wavelength of a 1 MHz signal is 300 m. A quarter wavelength, the length of one side of a dipole antenna, would be 75 m. Call it 72 m because 72 has lots of small factors, i.e. it’s usually mentally easier to divide things into 72 than 75. Rounding 75 down to 72 results in the antenna being a little shorter than ideal. But antennas are forgiving, especially for receiving.

Update: There’s more to replacing 75 with 72 than simplifying mental arithmetic. See Markus Meier’s comment below.

Just as the Rule of 72 for antennas rounds 75 down to 72, the Rule of 72 for interest rounds 69.3 up to 72, both for ease of mental calculation.

$\begin{align*} \left(1 + \frac{n}{100}\right)^{72/n} &= \exp\left(\frac{72}{n} \log\left(1 + \frac{n}{100}\right)\right) \\ &\approx \exp\left(\frac{72}{n} \, \frac{n}{100} \right) \\ &= \exp(72/100) \\ &= 2.05 \end{align*}$

The approximation step comes from the approximation log(1 + x) ≈ x for small x, a first order Taylor approximation.

The last line would be 2 rather than 2.05 if we replaced 72 with 100 log(2) = 69.3. That’s where the factor of 69.3 mentioned above comes from.

[1] The post actually says centimeters, but the author meant to say meters.

Handy approximation for roots of fractions

Posted on 24 February 2024 by John

This post will discuss a curious approximation with a curious history.

Approximation

Let x be a number near 1, written as a fraction

x = p / q.

Then define s and d as the sum and difference of the numerator and denominator.

s = p + q
d = p − q

Since we are assuming x is near 1, s is larger relative to d.

We have the following approximation for the nth root of x.

ⁿ√x ≈ (ns + d) / (ns − d).

This comes from a paper written in 1897 [1]. At the time there was great interest in approximations that are easy to carry out by hand, and this formula would have been very convenient.

The approximation assumes x is near 1. If not, you can multiply by a number of known square root to make x near 1. There will be an example below.

Examples

Positive d

Let’s find the cube root of x = 112/97. We have n = 3, p = 112, q = 97, s = 209, and d = 15. The approximation tells says

³√x ≈ 642/612 = 107/102 = 1.049019…

while the exact value is 1.049096… .

Negative d

The value of d might be negative, as when x = 31/32. If we want to find the fifth root, n = 5, p = 31, q = 32, s = 63, and d = −1.

⁵√x ≈ 312/314= 156/157 = 0.9936708…

while the exact value is 0.9936703… .

x not near 1

If x is not near 1, you can make it near 1. For example, suppose you wanted to compute the square root of 3. Since 17² = 289, 300/289 is near 1. You could find the square root of 300/289, then multiply the result by 17/10 to get an approximation to √3.

History

The author refers to this approximation as Mercator’s formula, presumable Gerardus Mercator (1512–1594) [2] of map projection fame. A brief search did not find this formula because Mercator’s projection drowns out Mercator’s formula in search results.

The author says a proof is given in Hutton’s Tracts on Mathematics, Vol 1. I tracked down this reference, and the full title in all its 19th century charm is

TRACTS
ON
MATHEMATICAL
AND
PHILOSOPHICAL SUBJECTS,
COMPRISING,
AMONG NUMEROUS IMPORTANT ARTICLES,
THE THEORY OF BRIDGES,
WITH SEVERAL PLANS OF IMPROVEMENT,
ALSO,
THE RESULTS OF NUMEROUS EXPERIMENTS ON
THE FORCE OF GUNPOWER,
WITH APPLICATIONS TO
THE MODERN PRACTICE OF ARTILLERY.
IN THREE VOLUMES
BY CHARLES HUTTON, LL.D. AND F.R.S. &c.
Late Professor of Mathematics in the Royal Military Academy, Woolwich.

Hutton’s book looks interesting. You can find it on Archive.org. Besides bridges and gunpowder, the book has a lot to say about what we’d now call numerical analysis, such as ways to accelerate the convergence of series. Hutton’s version of the formula above does not require that x be near 1.

[1] Ansel N. Kellogg. Empirical formulæ; for Approximate Computation. The American Mathematical Monthly. February 1897, Vol. 4 No. 2, pp. 39–49.

[2] Mercator’s projection is so familiar that we may not appreciate what a clever man he was. We can derive his projection now using calculus and logarithms, but Mercator developed it before Napier developed logarithms or Newton developed calculus. More on that here.

Doomsday 2024

Posted on 26 December 2023 by John

John Conway’s “Doomsday” rule observes that that every year, the dates 4/4, 6/6, 8/8, 10/10, 12/12, 5/9, 9/5, 7/11, and 11/7 fall on the same day of the week. In 2024 all these dates fall on a Thursday.

These dates are easy to memorize because they break down in double pairs of even digits—4/4, 6/6, 8/8, 10/10, and 12/12— and symmetric pairs of odd digits—5/9 and 9/5, 7/11 and 11/7. Because of their symmetry, this set of dates is the same whether interpreted according to the American or European convention.

In addition, the last day of February falls on “Doomsday.” Since 2024 is a leap year, this will be February 29.

Related post: Mentally calculating the day of the week

Mentally approximating factorials

Posted on 22 June 2023 by John

Suppose you’d like to have a very rough idea how large n! is for, say, n less than 100.

If you’re familiar with such things, your Pavlovian response to “factorial” and “approximation” is Stirling’s approximation. Although Stirling’s approximation is extremely useful—I probably use it every few weeks—it is not conducive to mental calculation.

The cut point

It’s useful to know that 24! ≈ 10²⁴ and 25! ≈ 10²⁵.

Said another way, the curves for n! and 10ⁿ cross approximately midway between 24 and 25. To the left of the crossing, n! < 10ⁿ and to the right of the crossing n! > 10ⁿ.

So, for example, if you hear someone refer to permutations of the English alphabet, you know the number of permutations is going to be north of 10²⁶.

Left of the cut

Suppose you want to estimate n! for n < 24. You know n! < 10ⁿ, but maybe you’d like to be a little more precise.

I’ll suppose you know n! for n up to 6. The approximation

log₁₀ n! ≈ n − 2

has an absolute error of less than 1.5 for n = 7, 8, 9, …, 23.

Right of cut

For n = 26, 27, 28, …, 100 the approximation

log₁₀ n! ≈ 7n/4 − 20

has an absolute error less than 3.

Note that calculating 7n/4 as n + n/2 + n/4 is probably easier than calculating (7n)/4.

Simple approximation

Origin story

Related posts

Dates in the 20th century

Dates in future centuries

Why the rules work

Partial explanation

Full explanation

Related posts

Divisibility and factoring

Day of the week

Transcendental functions

Logarithms

Miscellaneous

Related posts

Related posts

Simplest approach

First refinement

Second refinement

Related posts

Related posts

Approximation

Examples

Positive d

Negative d

x not near 1

History

Related posts

The cut point

Left of the cut

Right of cut

Related posts