Can you find a pattern in this?

With this equation, is there some sort of pattern that leads to the points that work?

x^y=y^x

I mean, I know that if x and y are the same, that the answer will be the same, but there are some others that work(2,4)...So is there some kind of pattern to this?

Only thing I can think of is to use a logarithm...

Ln(x^y)=Ln(y^x)
y*Ln(x)=x*Ln(y)
y/x = Ln(y)/Ln(x)

If that helps to get any more results... It's the only way I can think of to simplify/change it...

Quote:

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Originally Posted by manavo11

Only thing I can think of is to use a logarithm...

Ln(x^y)=Ln(y^x)
y*Ln(x)=x*Ln(y)
y/x = Ln(y)/Ln(x)

If that helps to get any more results... It's the only way I can think of to simplify/change it...

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Thank you, I'm thinking that their is no pattern, and their probably isn't. I made a program that gave me all the coordinates that worked up to a certain point, and they all looked completely random. I just saw this problem on another website, and wondered if their was some kind of pattern, but I guess their's not.

Looks to me that,
for a specific M, there exists an A;
A = M^(1/(M-1))

And from this you can get a B:
B = AM

which A and B then satisfies:

A^B= B^A



*************************************
So for example, if we want to find A, Given M = 2:

A = 2^(1/(2-1)); or A = 2.

Thus, B = A*M, or 4,
and we see that
A^B = B^A is valid, or 2^4 = 4^2

***********Or, Another Example:******************
Lets say M = 7, that makes:
A = 1.3830875542684884926406585135348

And then:
B = 9.6816128798794194484846095947438

So, does A^B = B^A?

A^B = 23.102129367646701645408458653
B^A = 23.102129367646701645408458653

Yep!

:wave:

Quote:

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Originally Posted by NotLKH

Looks to me that,
for a specific M, there exists an A;
A = M^(1/(M-1))

And from this you can get a B:
B = AM

which A and B then satisfies:

A^B= B^A



*************************************
So for example, if we want to find A, Given M = 2:

A = 2^(1/(2-1)); or A = 2.

Thus, B = A*M, or 4,
and we see that
A^B = B^A is valid, or 2^4 = 4^2

***********Or, Another Example:******************
Lets say M = 7, that makes:
A = 1.3830875542684884926406585135348

And then:
B = 9.6816128798794194484846095947438

So, does A^B = B^A?

A^B = 23.102129367646701645408458653
B^A = 23.102129367646701645408458653

Yep!

:wave:

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Wow! You are extremely smart, I know I couldn't do that. Thank you very much! :)

OOPS!
Sorry, I realized that I didn't show how I got the A M Relationship.

Lets say, for some A,B, that:
A^B = B^A

Certainly, then, for some M,
A^M= B

And for some R:
(A^M)^R = B^R

Or:
A^RM=B^R

So, if the two equations are identical, or:
A^RM = B^R <==> A^B = B^A

Then, equivalently:
RM = B and A = R.
Or...
AM = B

Now, Going back to the equation A^M= B, we see that
A^M= AM

Divide both sides by A, or actually multply both sides by A^-1:
A^M-1=M

Now, take the M-1 root of both sides:

(A^M-1)^(1/(M-1)) = M^(1/(M-1))

Which then gives us, for some M:
A = M^(1/(M-1))

Thus, for some M, we have an A, B such that A^B=B^A when:

A = M^(1/(M-1))
B = AM = M*[M^(1/(M-1))] = M^(M/(M-1))

:)

Quote:

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Originally Posted by System_Error

Wow! You are extremely smart, I know I couldn't do that. Thank you very much! :)

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:thumb:

Hmmm,
Here's a neet way to say the same thing:

If we say A = (1/a)
And B = (M/a)

it can be worked out that M*(a^M-1) = 1.

So, if M = 2, 3, 4, ...
Then a = (1/2), sqrt(1/3), cuberoot(1/4), ...

:wave: