Here's all you really need to know about logs when estimating in your head:
The number of digits minus one is the magnitude (integer). Then add the leading digit like so:
1x = ^0.0
2x = ^0.3 (actually ^0.301...)
pi = ^0.5 (actually ^0.497...)
5x = ^0.7 (actually ^0.699...)
Between these, you can interpolate linearly and it's fine for estimating. Also 3x is close enough to pi to also be considered ^0.5.
In fact, if all you're doing is estimating, you don't even really need to know the above log table. Just use the first digit of the original number as the first digit past the decimal. So like 6000 would be ^3.6 (whereas it's actually ^3.78). It's "wrong" but not that far off if you're using logarithmetic for napkin math.
xeonmc · 1d ago
And this is also the basis of the fast inverse square root algorithm. Floating point numbers are just linear interpolations between octaves.
I find much more useful to memorize the inverse of that table, i.e. the decibel table.
The numbers corresponding approximately to 0 dB, 1 dB, 2 dB, 3 dB etc. (1.0, 1.25, 1.6, 2.0 etc.; more accurate values, like 1.26 or 1.58 instead of 1.25 and 1.6 are typically not needed) are also frequently encountered as such in engineering practice as belonging to various series of standard nominal values, which makes them more useful to have in mind.
Then for any mental computation, it is trivial to convert mentally a given number to the corresponding dB value, perform the computation with additions and subtractions instead of multiplications and divisions, then convert back the dB value to the linear number corresponding to the result.
Doing mental computations in this way, even if it has only about 2 correct digits at most, is enough for debugging various hardware problems, or even for catching software bugs, where frequently an impressive numbers of significant digits is displayed, but the order of magnitude can be completely wrong.
madcaptenor · 15h ago
Since 3 dB is so nearly 2.0, you can derive that series by taking small powers of two, say from 2^(-4) to 2^5 or 2^6:
The pi = ^0.5 bit reminds me of a college physics professor who was fond of using the shortcut π²=10.
madcaptenor · 1d ago
This is wrong, π² = g, the acceleration due to gravity.
thomasahle · 1d ago
What is this ^notation?
Looks like 5x=^0.699 means log_10(5)=0.699.
saulpw · 1d ago
It's magnitude notation. ^X is short for 10^X.
xeonmc · 1d ago
5 = 10^0.699
thechao · 1d ago
I don't know about powers-of-10; but, you can use something similar to bootstrap logs-in-your-head.
So, 2^10=1024. That means log10(2)~3/10=0.3. By log laws: 1 - .3 = 0.7 ~ log10(5).
Similarly, log10(3)*9 ~ 4 + log10(2); so, log10(3) ~ .477.
Other prime numbers use similar "easy power rules".
Now, what's log10(80)? It's .3*3 + 1 ~ 1.9. (The real value is 1.903...).
The log10(75) ~ .7*2+.477 = 1.877 (the real answer is 1.875...).
Just knowing some basic "small prime" logs lets you rapidly calculate logs in your head.
madcaptenor · 1d ago
For log(3) I prefer the "musical" approximation 2^19 ~ 3^12. This is a "musical" fact because it translates into 2^(7/12) ~ 3/2 - that is, seven semitones make a perfect fifth). Together with log(2) ~ 3/10 that gives log(3) ~ 19/40.
Also easy to remember: 7^4 = 2401 ~ 2400. log(2400) = log(3) + 3 log(2) + 2 ~ 19/40 + 3 * 12/40 + 2 = 135/40, so you get log(7) ~ 135/160 = 27/32 = 0.84375.
thechao · 15h ago
I'll double-reply, since I think people might also appreciate this... there's a pretty easy way to find the square of 2-leading-digit numbers (and square-roots, too.)
Direct square-root: 3800 = (61^2=3721) + r ~ (61 + (3800-3721)/(2*61) = 61 + 79/122. To check: (61 + 79/122)^2= 3,800.419. You can estimate the "overhead" by noting that 79/122 ~ 2/3 => (2/3)^2 = 0.444.
Of course, my grandfather would've just used a slide-rule directly.
thechao · 1d ago
These are both great! I learned most of these old tricks from my dad & grandfather.
xeonmc · 1d ago
Protip: since halving and doubling are the same logarithmic distance on either sides of unity, and the logarithmic distance of 2.0 to 5.0 is just a tiny bit larger than that of doubling, this means that you can roughly eyeball the infra-decade fraction by cutting them into thirds
teo_zero · 22h ago
The artcle makes a weird use of the notation for successive exponentiations. It's difficult to represent the concept here as only text is allowed, but shortly the author uses a^b^c to mean (a^b)^c. This is counter intuitive as in most mathematical contexts exponentiation is right-associative, that is a^b^c means a^(b^c). A pair of parentheses would remove any doubts!
(I see that someone already mentioned fast inverse square root algorithm is related to this, which is famously used by John Carmack which is one of my hero who led me into tech industry, despite I didn't end up in gaming industry)
briian · 1d ago
So much of economics maths/stats is built on this one little trick.
It's still pretty cool to me that A this works and B it can be used to do so much.
gus_massa · 1d ago
Typo near the top, in case someone knows the author:
> log(100)≤log(N)<log(100)
There is a missing 0 in the last log. It should be
The number of digits minus one is the magnitude (integer). Then add the leading digit like so:
1x = ^0.0
2x = ^0.3 (actually ^0.301...)
pi = ^0.5 (actually ^0.497...)
5x = ^0.7 (actually ^0.699...)
Between these, you can interpolate linearly and it's fine for estimating. Also 3x is close enough to pi to also be considered ^0.5.
In fact, if all you're doing is estimating, you don't even really need to know the above log table. Just use the first digit of the original number as the first digit past the decimal. So like 6000 would be ^3.6 (whereas it's actually ^3.78). It's "wrong" but not that far off if you're using logarithmetic for napkin math.
So you only need to remember:
The numbers corresponding approximately to 0 dB, 1 dB, 2 dB, 3 dB etc. (1.0, 1.25, 1.6, 2.0 etc.; more accurate values, like 1.26 or 1.58 instead of 1.25 and 1.6 are typically not needed) are also frequently encountered as such in engineering practice as belonging to various series of standard nominal values, which makes them more useful to have in mind.
Then for any mental computation, it is trivial to convert mentally a given number to the corresponding dB value, perform the computation with additions and subtractions instead of multiplications and divisions, then convert back the dB value to the linear number corresponding to the result.
Doing mental computations in this way, even if it has only about 2 correct digits at most, is enough for debugging various hardware problems, or even for catching software bugs, where frequently an impressive numbers of significant digits is displayed, but the order of magnitude can be completely wrong.
0.0625, 0.125, 0.25, 0.5, 1, 2, 4, 8, 16, 32, (64)
and then multiplying everything by factors of 10 to get it between 0 and 10:
1, 1.25, 1.6, 2, 2.5, 3.2, 4, 5, 6.25 (or 6.4), 8
Same for 9: sqrt(81) = sqrt(8 * 10) -> (0.3*3 + 1) / 2 = 0.95
At some point it's easier to remember a few numbers than calculate a lot of formulas.
Put the effort into remembering a 3 digit log for 7 instead?
Or keep the same precision with ...
Log 2 is all you need?Or even ...
VERY accurate, but that's getting to be too much.Looks like 5x=^0.699 means log_10(5)=0.699.
So, 2^10=1024. That means log10(2)~3/10=0.3. By log laws: 1 - .3 = 0.7 ~ log10(5).
Similarly, log10(3)*9 ~ 4 + log10(2); so, log10(3) ~ .477.
Other prime numbers use similar "easy power rules".
Now, what's log10(80)? It's .3*3 + 1 ~ 1.9. (The real value is 1.903...).
The log10(75) ~ .7*2+.477 = 1.877 (the real answer is 1.875...).
Just knowing some basic "small prime" logs lets you rapidly calculate logs in your head.
Also easy to remember: 7^4 = 2401 ~ 2400. log(2400) = log(3) + 3 log(2) + 2 ~ 19/40 + 3 * 12/40 + 2 = 135/40, so you get log(7) ~ 135/160 = 27/32 = 0.84375.
Direct square: 54^2 = (50 + 4)^2 = 2500 + 2 * 50 * 4 + 4^2.
Direct square-root: 3800 = (61^2=3721) + r ~ (61 + (3800-3721)/(2*61) = 61 + 79/122. To check: (61 + 79/122)^2= 3,800.419. You can estimate the "overhead" by noting that 79/122 ~ 2/3 => (2/3)^2 = 0.444.
Of course, my grandfather would've just used a slide-rule directly.
Also related: https://blog.timhutt.co.uk/fast-inverse-square-root/
(I see that someone already mentioned fast inverse square root algorithm is related to this, which is famously used by John Carmack which is one of my hero who led me into tech industry, despite I didn't end up in gaming industry)
It's still pretty cool to me that A this works and B it can be used to do so much.
> log(100)≤log(N)<log(100)
There is a missing 0 in the last log. It should be
> log(100)≤log(N)<log(1000)