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        riddles >> easy >> power ranges
        
(Message started by: Noke Lieu on Mar 20^th, 2012, 3:06pm)

Title: power ranges
Post by Noke Lieu on Mar 20^th, 2012, 3:06pm

If A and B are positive integers, and a<b, does it follow that a^b < b^a ? When?

Title: Re: power ranges
Post by towr on Mar 20^th, 2012, 10:59pm

[hide]2^3 < 3^2
2^4 = 4^2
2^5 > 5^2[/hide]

Title: Re: power ranges
Post by pex on Mar 21^st, 2012, 1:46am

on 03/20/12 at 15:06:29, Noke Lieu wrote:

If A and B are positive integers, and a<b, does it follow that a^b < b^a ? When?

In fact, it is "usually" the other way around. Your statement holds if [hide]a=1, and if (a,b)=(2,3), but not in any other cases. It is not a coincidence that things change around the number e ;)[/hide]

Title: Re: power ranges
Post by Noke Lieu on Mar 21^st, 2012, 10:09pm

Yeah- I wound up at when a/b > log_ba

I was hoping that it'd be more enlightening than that. I guess pex, you make a nice point in that vein.

I supose I shouldn't have restricted it to integers, but even then...

Title: Re: power ranges
Post by pex on Mar 22^nd, 2012, 12:38am

on 03/21/12 at 22:09:01, Noke Lieu wrote:

Yeah- I wound up at when a/b > log_ba

... which is equivalent to (log a)/a < (log b)/b. And (log x)/x is maximized at x=e.

Over the integers, (log 3)/3 is the maximum. Further, (log 2)/2 = (log 4)/4. Finally, (log 1)/1 = 0 and (log n)/n is positive for all other integers. My answer was derived from these properties.

So your puzzle was more interesting than you think :)