[Guest]

Given that x and y are positive integers, solve x^y = y^x.

*Note: x and y are unique. i.e. x!=y

Edited June 4, 2009 by adiace

[Guest]

X = Y. the can be any positive interger, as long as they are both the same one

[Guest]

Oops, just read the note, disregard my previous post

[Guest]

x=2 y=4

[CaptainEd]

edit: my answer was eclipsed by shade

Edited June 4, 2009 by CaptainEd

[Guest]

x=2

y=4

4x4=16

2x2x2x2=16

Dang it. I actually got one and some one beat me to the punch

Edited June 4, 2009 by scraner

[Guest]

edit: my answer was eclipsed by shade

How do you know there is no other solution?

[Guest]

Here is a separate answer which I think satisfies the question.

X=1

Y=2i

[Guest]

Here is a separate answer which I think satisfies the question.

X=1

Y=2i

Lets keep it real

* positive integer

[Guest]

I'd say if x=1, y could equal any positive number.

x=1, y=273 1^273=273^1

[Guest]

I'd say if x=1, y could equal any positive number.

x=1, y=273 1^273=273^1

1^273 does not equal 273^1

1^273 =1

273^1 = 273

[Guest]

I'd say if x=1, y could equal any positive number.

x=1, y=273 1^273=273^1

But 1 times 1 is always 1, where as any integer one time is always the same integer. This does not work.

Example, same as shown:

1x1x1x1x1x1x1x1x1...(273 times) = 1

273 = 273 (1 time)

Oh, Shade beat me!

Edited June 4, 2009 by natille

[Guest]

x=4

y=2

[Guest]

instead of X, how about let Y=1 and X be any other positive integer...

there was no stipulation on V, so let V=0...

Edited June 4, 2009 by icepick1346

[Guest]

2,4

[Guest]

Lets keep it real

* positive integer

Any other restrictions?

[Guest]

@ those who arrived at 2 and 4.

Can you prove 2, 4 is the only possible solution?

[Guest]

delete this, <-- wasn't thinking.

Edited June 4, 2009 by Azy

[Guest]

1^273 does not equal 273^1

1^273 =1

273^1 = 273

OOPs maybe should have thought more about that.

[Guest]

@ those who arrived at 2 and 4.

Can you prove 2, 4 is the only possible solution?

could'nt you add a 0 to each side, like 20, 40 , 200,400....?

[Guest]

1^(-1)=1

-1^(1)=-1

doesn't work

*they fixed their post

Edited June 4, 2009 by perk8504

[Guest]

could'nt you add a 0 to each side, like 20, 40 , 200,400....?

nevermind, just tried it, don't work

[Guest]

just a note to make: x^(y/x) must equal y as well, might help in finding the pattern

take the log of both sides and rearrange

[Guest]

with that...2 and 4 are the only positive integers that fit the restraints. because that equation says that one number must be the square of the other, and no other squares fit

*also, any positve and negative combination of 2s and 4s works

Edited June 4, 2009 by perk8504

[Guest]

with that...2 and 4 are the only positive integers that fit the restraints. because that equation says that one number must be the square of the other, and no other squares fit

*also, any positve and negative combination of 2s and 4s works

Not true... there is another set of positive integers that satisfies the stipulations of the puzzle. However, I don't think the OP likes my answer... he is too grounded in reality.

[Guest]

Not true... there is another set of positive integers that satisfies the stipulations of the puzzle. However, I don't think the OP likes my answer... he is too grounded in reality.

Just nitpicking with this, but I'm pretty sure -1^(1/2) is not considered positive, and then curious, is it actually considered an integer?