3 ms·
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
by thechao 1y 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 1y agoFor 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 1y agoThese are both great! I learned most of these old tricks from my dad & grandfather.
- thechao 1y agoI'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.