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Estimating Logarithms

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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.

And this is also the basis of the fast inverse square root algorithm. Floating point numbers are just linear interpolations between octaves.

What is this ^notation?

Looks like 5x=^0.699 means log_10(5)=0.699.

5 = 10^0.699

It's magnitude notation. ^X is short for 10^X.

The pi = ^0.5 bit reminds me of a college physics professor who was fond of using the shortcut π²=10.

This is wrong, π² = g, the acceleration due to gravity.

Slightly more accurate ..

    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)

The biggest error is 7 at +0.577% ... 0.845 is almost perfect. The others are maximum +0.45% off.

So you only need to remember:

    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

You barely even have to remember

  9: 0.95

since you can get it by interpolation between 8 and 10.

Yup.

Put the effort into remembering a 3 digit log for 7 instead?

Or keep the same precision with ...

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

Log 2 is all you need?

Or even ...

    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

VERY accurate, but that's getting to be too much.

You also don't need to remember 7, as 2*7^2 =~ 100 => log 7 =~ (2 - 0.3) / 2 = 0.85

Yes, as I mentioned in one of my other replies. Or 2400^0.25.

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.

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.

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:

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

You are right, so this is how the series of standardized nominal values that has been traditionally used for some quantities, like maximum dissipated powers for resistors, maximum operating voltages for capacitors, maximum powers for motors, hexagonal nut sizes for some ancient packages for stud-mounted power semiconductor devices, etc., has been obtained.

Apparently whoever have standardized first this series of values have not been able to make their mind to choose between 6.25 and 6.4, therefore 6.3 has been chosen instead of any of those 2 values.

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.

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.

These are both great! I learned most of these old tricks from my dad & grandfather.

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: 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.

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

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!

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

(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)

Thanks for pointing this out :), I fixed it!

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