3 ms·
I did it similarly: log(19,683) is about 4.3, so ln(19,683) is about 4.3 * 2.3, which is about 9.9. That jumps out as close to 9ln(3). So the natural log of th
by colinbeveridge 4y ago
I did it similarly: log(19,683) is about 4.3, so ln(19,683) is about 4.3 * 2.3, which is about 9.9. That jumps out as close to 9ln(3).
So the natural log of the cube root is (presumably) 3 ln(3), making the cube root itself 27.
Alternatively, the digit sum of 19,683 is 27, so the cube is a multiple of 9. Given that it's a perfect cube, 19,683 must be a multiple of 3^3 = 27. Dividing by that gives 729, or 9^3, so the cube root is 3 * 9=27.
Lastly, it's possible to dredge up that e^3 is about 20, so (27.1828...)^3 is about 20,000, and it's plausible that 27 is the cube root we're after.
- cperciva 4y agoMy approach: ln(19683) ~ ln(20000) = ln(20) + 3 ln(10) ~ 3 + 3*2.3 = 9.9. 3.3 = 1.0 + 2.3 ~ ln(e) + ln(10). My mental landmarks are ln(2), ln(e), and ln(10), roughly 0.7, 1.0, and 2.3 respectively.
- colinbeveridge 4y agoNice. I like ln(3) and ln(5) (1.1 and 1.6) as well.
- cperciva 4y agoYeah, ln(3) is useful, although you can also get it from ln(1+x) ~ x. I don't bother remembering ln(5) though; it's just ln(10) - ln(2) anyway.