x^y=y^x

x,y are whole numbers (Z).

Can you solve it?

x and y are any two integers such that x=y.

How exactly did you get to the answer (mathematically)?

2^4 = 4^2

My first reaction is x=y.
However, a more unified answer is harder to find.
this is developing as I type... so maybe I'll make a mistake or get stuck (forgive me if I do :rolleyes: )
let y=x+a
x^y = y^x <=>

x^(x+a) = (x+a)^x <=>
x^x + x^a = x^x + x*ax^(x-1) + ...+x*xa^(x-1)+a^x (Pascal's sequence) <=>
x^a = x*ax^(x-1)+ 2(x-1)*a^2*x^(x-2)+...+2(x-1)*a(x-2)*x^2+x*xa^(x-1)+a^x <=>
some cleaning up to do...
x^a = ax^x + ... Hmmm I need a paper now. I'll get back to this later.

Merlijn, you're a genious.

does this mean you have the answer now?
or do I have to continue?
(or were you merely sarcastic?) :)

BloodSuckingGerbile:

What's <b>your</b> answer to this?

No........
I don't have the answer... yet.
But you gave me a starting point.
continue, of course!
Let's see who gets first to the answer.

James R:
I don't know the answer. Trying to find it, though.

I took Marlijns' example and tried doing what he did but it seemed too much to "clean up" so I assumed y=ax

and I got

x^(ax)=(xa)^x
||
\/
(x^x)^a=x^x*a^x
||
\/
(x^x)^(a-1)=a^x

And then I did

ln[(x^x)^(a-1)]=ln(a^x)
|| ||
(a-1)*x*lnx = x*lna
||
\/
lnx=(lna)/(a-1)

and then

e^(lnx)=e^[(lna)/a-1)]
|| ||
x = (a-1)st root of a = a^[1/(a-1)]

try it on a calculator. It actually works. Now, this is true for real numbers, but I need to find all a's for which the expression

a^[1/(a-1)] = Z

is true.

Merlijn, try to finish what you started. Maybe that will give the answer.

BloodSuckingGerbile,
Ok, I will try to work on it; it's a lot of cleaning up though. Maybe there is an easier way.
However, I seriously doubt that y=ax makes it any easier.

bye

I think I have the answer.

(a-1)*x*lnx = x*lna

(before I divided lna by a-1)
Gives the answer a=1, or y=x*1=x, which is true.
Then when I divided lna by (a-1)
I got

lnx=(lna)/(a-1)

and then

x=a^[1/(a-1)]

So there is only one integer a for which a^[1/(a-1)] is an integer and that is a=2, which gives x=2 and y=a*x=2*2=4.

:bugeye:

:eek: Just one more thing: -2 and -4 are also true...

%@#$%!!!!!! :mad:

I guess I'll have to do that from the beginning...

well you know that with your
y=ax
there is a solution a=1 (for all x) and a=2 (at least for x=2)
negative solutions for a are higly improbable (hmmm x=1, a=-1 gives 1/1 = -1 ? eeeh obviously not! :))
Forget the negative solutions.

But the negative solutions exist.
Merlijn, you think that with y=x+a you can get them too?

well only if both x and y are negative... but not only one of the two...
I haven't had time yet to have another look at the problem... be patient pleace.
(I am sure there are others who can solve the problem)

x=2
y=4

Cool problem, gerbil :)

I think your approach works well, but you have to take care. As such:

y = ax, x^y = y^x

x^(ax) = (ax)^x

(x^a)^x = (xa)^x

But this is where we have to pay attention. If x is odd, then we can just raise both sides to power 1/x to obtain:

x^a = xa, x odd

If x is even then we have two possibilities:

x^a = xa, and x^a = -xa.

In either case, if a < 0 then |1/x^|a|| = |xa| => 1/|a| = |x^(|a|+1)|. Since x and a are integers, 1/|a| must be an integer which can only be if |a|=1, but a is even. Therefore a>0. Note that this applies regardless of whether x is odd or even (which means y and x must have the same sign.)

In the second case, let's consider the possible values of a:

a=2: x^2 = -2x => x = -2, y = -4
a=4: x^4 = -4x => x^3 = -4 => -1 < x < -2
a=6: x^6 = -6x => x^5 = -6 => -1 < x < -2
...
as a approaches infinity, x approaches -1 but is never again an integer. Which means the only solution for this case is x=-2,y=-4.

Now let's return to the other case:

x^a = xa, a>0

So let's try some values of a:

a=1: x = x => y = x
a=2: x^2 = 2x => x = 2, y = 4
a=3: x^3 = 3x => x^2 = 3 => 1 < x < 2
a=4: x^4 = 4x => x^3 = 4 => 1 < x < 2
...
In the limit, as a becomes infinite x approaches 1 but is never again an integer. Which means that all values of a greater than 2 are excluded.

So your solution is:

a=1: y=x, x any integer (positive or negative)
a=2: x=2,y=4 and x=-2,y=-4

That's it as far as integer solutions go, unless I missed something...

Overdoze, thanks for fully answering my question.

Not being careful is my biggest problem. :o

I was just playing around with Mathematica to see what it can do.
It came up with the solution:

-x * ProductLog[ -Log[x]/x ]
________________________
Log[x]

This is an interesting function: for 0<=x<=e it evaluates to x, at e it has its maximum, after which it decreases and has limit 1. So the only Integer solutions this function gives at y=2 are x=2 and x=4. Other integer solutions are x=y=0 and x=y=1.
GRAPH (http://www.xs4all.nl/~baumer/graph.pdf)

Why Mathematica misses the obvious x=y solution is a mystery to me. Maybe I don't know the program well enough.

hmm...
That sounds interesting.
I've never heard of Mathematica.
Some kind of software?
Tell me more, please.

Mathematica is the program from Wolfram Research that can do symbolic mathematics. See http://www.wolfram.com

Greetings,

Han.

What's ProductLog?

Sorry, I thought I explained, but obviously I didn't.
ProductLog is the inverse function of x*e^x, so ProductLog(p) gives you the number x such that x*e^x = p.

See: http://functions.wolfram.com/ElementaryFunctions/ProductLog2/

Greetings,


Han.

I wish I had Mathematica, but cannot afford it.

I have MathCad 7 and PSI Plot 5, both of which will do surface plots, as will Mathematica. I never use these capabilities and do not want to spend the time now on this problem.

Perhaps a Mathematica owner might want to use the surface plot capabilities to check for all real solutions.

Superimpose the following three surface plots.

z = x^y

z = y^x

The plane y = x

Each surface intersects the plane at possible solutions. If both surfaces intersect at the same height off the XY-Plane, you have a solution.

It should not be difficult to pick out possible solutions, verify them numerically, and be pretty sure you found them all.

BTW: Has anybody considered complex solutions? I have not studied all the posts carefully. Perhaps this has already been done. If not, it is an interesting extension of the problem.

BTW: Has anybody considered complex solutions? I have not studied all the posts carefully. Perhaps this has already been done. If not, it is an interesting extension of the problem.
Well, x,y,z, are N, but it really is interesting...

OKAY EVERYBODY!!

UPDATE!!!!

X,Y and Z ARE COMPLEX!!!

Thanx, Dinosaur :)

you can't use complex numbers, as given in the problem statement


anyways, the formal solution is anything that satisfies the following system


k=1+ ln(k)/ln(x) ;
where
y=k*x; and y and x are non-zero integers (hence is k) , as that is a trivial solution space anyways

given that this system is underspecified, only iterative solution seeking can work, as evidenced earlier

Is x^y defined for every x and y? According to my math-book (Thomas M. Apostol) x^y is defined to be exp(y*Log(x)) for x,y>0. The Exp and Log funtion are defined for complex numbers, so maybe x^y can be defined for complex x and y. Does anyone know this?


Greetings,


Han.

Originally posted by Han Baumer
Is x^y defined for every x and y? According to my math-book (Thomas M. Apostol) x^y is defined to be exp(y*Log(x)) for x,y>0. The Exp and Log funtion are defined for complex numbers, so maybe x^y can be defined for complex x and y. Does anyone know this?


Greetings,


Han.


duh.

Originally posted by ChristCrusher



duh.

I take it you are a real Baevis&Butthead fan!

i take it you don't realize your question answered itself?

http://lamer.net/pimplsap.gif

Originally posted by ChristCrusher
i take it you don't realize your question answered itself?

http://lamer.net/pimplsap.gif

Correct! I think you mean that x^y is defined to be Exp(y*Log(x)) for Complex x and y where x,y <>0. I suspected so, but I couldn't get any confirmation.

Well, if that is the case the solution to the problem x^y = y^x is as follows:

x^y = y^x
= { definition of x^y, x, y<>0}
Exp(y*Log(x)) = Exp(x*Log(y))
= { Exp(x) = Exp(y) <=> x=y }
y*Log(x) = x*Log(y)
= { x,y <>0, division by x and by y}
1/x*Log(x) = 1/y*Log(y)
= { Log(x) = -Log(1/x)}
1/x*Log(1/x) = 1/y*Log(1/y)
= { x = Exp(Log(x)) }
Exp(Log(1/x)))*Log(1/x) = 1/y*Log(1/y)
= { Definition of ProductLog: x*Exp(x) = y <=> x=ProductLog(y) }
Log(1/x) = ProductLog(1/y*Log(1/y))
= { Exp(x) = Exp(y) <=> x=y, x = Exp(Log(x)) }
1/x = Exp(ProductLog(1/y*Log(1/y)))
= { 1/x = Exp(y) <=> x=Exp(-y) }
x=Exp(-ProductLog(1/y*Log(1/y))) http://lamer.net/headbang.gif

There has to be a step in the proof where a '<=' should be instead of '='. I cannot find it. This has to be the case since the last formula doesn't describe all the solutions, since clearly x=y is also a solution.



Greetings,


Han.