### Brain Teasers

# Reversible Exponents

Math
Math brain teasers require computations to solve.

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?

### Hint

^ = to the power of. They are both single digits.### Answer

2 and 4. 2^4 = 16, and 4^2 = 16.Hide Hint Show Hint Hide Answer Show Answer

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## Comments

Like homework from Hades...

Agh, you know I can't do these! I don't even understand googol! Nice one.

Oct 07, 2006

2 and 4. By the way,how do I earn points?

Oct 07, 2006

I need 250 points! Help!

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!

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!

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

Oct 30, 2006

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

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

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

So, for each value of t with 0

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

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.

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):

y(x)=ProdLog[-Ln[x]/x]/(-Ln[x]/x)

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

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):

y(x)=ProdLog[-Ln[x]/x]/(-Ln[x]/x)

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

I got two solutions:

2^4 = 4^2 = 16

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

2^4 = 4^2 = 16

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

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