# x^y=y^x

Discussion in 'Physics & Math' started by BloodSuckingGerbile, Jul 5, 2002.

1. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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x^y=y^x

x,y are whole numbers (Z).

Can you solve it?

3. ### James RJust this guy, you know?Staff Member

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

5. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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

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2^4 = 4^2

8. ### Merlijncurious catRegistered Senior Member

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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

)
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.

9. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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Merlijn, you're a genious.

10. ### Merlijncurious catRegistered Senior Member

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

11. ### James RJust this guy, you know?Staff Member

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BloodSuckingGerbile:

12. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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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.

13. ### Merlijncurious catRegistered Senior Member

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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

14. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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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:

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

%@#\$%!!!!!!

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

15. ### Merlijncurious catRegistered Senior Member

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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.

16. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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

17. ### Merlijncurious catRegistered Senior Member

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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)

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x=2
y=4

19. ### overdozehumanRegistered Senior Member

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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.

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.

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...

20. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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Overdoze, thanks for fully answering my question.

Not being careful is my biggest problem.

21. ### Han BaumerMemberRegistered Senior Member

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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

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

22. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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hmm...
That sounds interesting.
I've never heard of Mathematica.
Some kind of software?