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Topic:
7^sqrt(8) > 8^sqrt(7) proof
Replies:
5
Last Post:
Jun 18, 1996 12:58 AM




Re: 7^sqrt(8) > 8^sqrt(7) proof
Posted:
Jun 13, 1996 11:35 AM


Here's a halfbaked idea I had for a calculus proof that 7^sqrt(8) > 8^sqrt(7).
(This problem is really starting to intrigue me because it seems so simple, but a clean and simple calculus proof has to date seemed rather elusive!)
Letting y stand for sqrt(56)/x, consider the function
f(x) = ln(x^y/y^x).
If we can show that f(sqrt(7)) > f(sqrt(8)), this would imply that
(sqrt(7)^sqrt(8))^2 > (sqrt(8)^sqrt(7))^2, which is equivalent to
7^sqrt(8) > 8^sqrt(7).
When I graph f(x) using Mathematica, it is visually unequivocal that
(*) f'(x) < 0 for sqrt(7) <= x <= sqrt(8)
SO  if someone can come up with a calculus proof of (*), that would, I think, complete a calculus proof of the original problem.
Interestingly, it's also visually clear that f'(x) changes sign somewhere inside [sqrt(7).2, sqrt(7)] and also somewhere inside [sqrt(8),sqrt(8 +.2].
Dan Asimov



