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Fortieth logs (Posted on 2011-05-31) Difficulty: 2 of 5
It turns out that the common logarithms of each of the numbers from 2 through 9 can be very well approximated by rational numbers of the form n/40.

Derive each of the numerators with no calculation aids beyond pencil and paper.

See The Solution Submitted by Jer    
Rating: 4.5000 (2 votes)

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Some Thoughts A start (spoiler) | Comment 1 of 15
2^10 = 1024
10* log 2 = log(2^10) = log(1024) ~ log(1000) = 3

==> log 2 ~ .3 = 12/40
==> log 4 = log 2 + log 2 ~ 24/40
==> log 8 = log 2 + log 4 ~ 36/40

Also,

3^4 = 81
4*log 3 = log 3^4 = log 81 = 1+log(8.1) ~ 1+log 8 ~76/40
==> log 3 ~ 19/40
==> log 6 = log 2 + log 3 ~ 31/40
==> log 9 = log 3 + log 3 ~ 38/40

That just leaves 5 and 7!

Edited on May 31, 2011, 2:29 pm
  Posted by Steve Herman on 2011-05-31 14:12:29

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