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Reversible Exponents

Math brain teasers require computations to solve.


Puzzle ID:#33340
Fun:*** (2.71)
Difficulty:** (1.75)
Submitted By:lessthanjake789*us******




There is only one pair of numbers that satisfies the equation x^y = y^x, where x does not equal y. Can you figure out which two numbers?

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Oct 05, 2006

Like homework from Hades...
Oct 05, 2006

Agh, you know I can't do these! I don't even understand googol! Nice one.
(user deleted)
Oct 07, 2006

2 and 4. By the way,how do I earn points?
(user deleted)
Oct 07, 2006

I need 250 points! Help!
Oct 09, 2006

quite easy, but fun nontheless.
the problem with these kind of teasers is that it has no real solution. it trial-and-error. so if you got the answers right, its always fun. but if you dont, its frustrating.
sometimes you just have to rely on luck and instinct.
its more challenging this way!
Oct 29, 2006

adeline, you earn one point for each time you submit a vote on a teaser. So vote on 250 of them and you will be sitting pretty
(user deleted)
Oct 30, 2006

have you proof that 2 and 4 are the only two distinct reversible exponents? what about -2 & -4?
Nov 23, 2006

Trial and error?
I disagree.

In few steps you can rewrite the equation as Ln[r]/r=t.
The function Ln[r]/r is only positive for r>1, and it has the maximum 1/e at r=e.
So, for each value of t with 0
Nov 23, 2006

So, for each value of t with 0
Nov 23, 2006

hmmm... it doesn't want to paste the whole answer... < is this character the problem?
Nov 23, 2006

So, for each value of t between 0 and 1/e, the above equation has 2 solutions, x and y, each of them in the intervals [1,e] and [e,Infinity) respectively. In other words, there are infinite solutions for this problem.
Nov 23, 2006

Assuming that the problem is restricted to the algebra of natural numbers, the only possible value for x between 1 and e is x=2 (e=2.718...). IF there is a solution at all, that solution is unique and it is with x=2. Replacing x=2 in the original equation gives y=4, and therefore the solution exist in the group of integer numbers.
Maybe the most challenging step is the last one, where you have to calculate one solution (let say y) as a function of the other one (x):


where w=ProdLog[z] is the well-known "logarithmic product" function (w is the solution of the equation z=w*e^w).
Jan 14, 2009

I got two solutions:

2^4 = 4^2 = 16
(-2)^(-4) = (-4)^(-2) = 0.0625

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