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

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 Daniel A. Asimov Posts: 101 Registered: 12/3/04
Re: 7^sqrt(8) > 8^sqrt(7) proof
Posted: Jun 13, 1996 11:35 AM

Here's a half-baked 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

Date Subject Author
6/13/96 Daniel A. Asimov
6/13/96 J. B. Rainsberger
6/14/96 Ronald Bruck
6/14/96 Daniel A. Asimov
6/18/96 David Shield