Estimating Logarithms

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.

    1: 0.0
    2: 0.3
    3: 0.475
    4: 0.6 (2*2)
    5: 0.7
    6: 0.775 (2*3)
    7: 0.85
    8: 0.9 (2*2*2)
    9: 0.95 (3*3)

    2: 0.3
    7: 0.85
    9: 0.95

    1.4 = sqrt(2) -> 0.15
    3 = sqrt(9) -> 0.95/2 = 0.475
    pi = ~sqrt(10) -> 0.5
    4 = 2*2 -> 0.6
    5 = 10/2 -> 0.7
    6 = 2*3
    2pi = 2*sqrt(10) -> 0.3+0.5 = 0.8
    8 = 2*2*2 -> 0.9

  9: 0.95

    7 = ~sqrt(100/2) -> (2-0.3)/2 = 1.7/2 = 0.85

    7 = sqrt(sqrt(2401)) = (3*8*100)^(1/4) -> 0.95/8 + 0.3*3/4 + 2/4 = 0.12 + 0.225 + 0.5 = 0.845

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.

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!

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

Related Wikipedia entry: https://en.wikipedia.org/wiki/Logarithmic_number_system

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)

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.

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

> log(100)≤log(N)<log(1000)