Wikipedia:Reference desk/Archives/Mathematics/2014 February 20
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February 20
editIs there a solution for 1×(1.5^a) = 1×(2^b) where A and B are positive integers?
editIs there a solution for 1×(1.5^a) = 1×(2^b) where A and B are integer?
If yes, what is the one with the smallest A — Preceding unsigned comment added by 201.78.133.48 (talk) 16:55, 20 February 2014 (UTC)
- Assuming the leading "1×" can be omitted from both sides, you're asking if 1.5a = 2b for any positive integers a,b.
- I don't believe 1.5n will ever be an integer (and there's probably a simple proof of that), so the answer is "no". Rojomoke (talk) 17:19, 20 February 2014 (UTC)
- 1.5 = 3/2, so a simple argument using simplest representation of a rational number shows that the only integer solution is a = b = 0 —Quondum 17:46, 20 February 2014 (UTC)
- Since the prime factorization of integers is unique, this should be true any time the two bases are unequal rational numbers. OldTimeNESter (talk) 21:47, 20 February 2014 (UTC)
- No. Consider the two unequal rational numbers 4/9 and 8/27. —Quondum 22:12, 20 February 2014 (UTC)
- Since the prime factorization of integers is unique, this should be true any time the two bases are unequal rational numbers. OldTimeNESter (talk) 21:47, 20 February 2014 (UTC)
- 1.5 = 3/2, so a simple argument using simplest representation of a rational number shows that the only integer solution is a = b = 0 —Quondum 17:46, 20 February 2014 (UTC)
- If there were a solution, we wouldn't need the Archimedian comma. --ColinFine (talk) 00:36, 26 February 2014 (UTC)