In one of Feynman's books he write about being able to do a lot of computations quickly by knowing just two common logarithms to three decimal places: log2=.301 and log3=.477
12^37 = (2*2*3)^37
Log(12^37)=37*(log2+log2+log3)=37*1.079=39.923
so the number is 10^39.923 = (10^0.923)*(10^39)
Focus on 10^0.923
log8=3log2=3*.301=.903
log9=2log3=2*.477=.954
Since our exponent between these, 10^0.923 is between 8 and 9, hence it begins with 8.
Mr. Feynman practiced this enough that he could do it in his head. I console myself that I did manage it with only P&P
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Posted by Jer
on 2020-11-18 11:04:21 |
P&P solution a'la Feynman
