Solution to y^x = x^y?

Question

Any ideas how to solve? (for x or y)

Answer 1

A previous answerer suggested x = 2 and y = 4 as solutions. These are correct, but there are infinitely many solutions, however those are the only rational solutions.

To avoid confusion with the x-y cord system, consider the problem
a^b = b^a
Raising both sides to the power 1/ab
a^(1/a) = b^(1/b)

Now this is the same thing as solving the following equation:
y = x^(1/x)
Are there values of x which give the same y value? As we've seen, x = 2 and x = 4 both give y = sqrt(2).

Let's investigate y = x^(1/x) some more.
ln(y) = ln(x) / x
1/y * dy/dx = (1 - lnx) / x^2
You'll find that
1) there is a maximum at x = e = 2.718...
2) as x --> 0, y --> 0 and dy/dx --> 0.
3) as x --> infinity, y --> 1, dy/dx --> 0.
4) y(1) = 1^1 = 1.
So for 1 < x < e, there is a corresponding x > e which yields the same y value.
eg x = 1.5 and x = 7.4088 give same value,
ie 1.5^7.4088 = 7.4088^1.5 = 20.166
x = 2.5 and x = 2.9703 give the same value
ie 2.5^2.9703 = 2.9703^2.5 = 15.205
etc
5) x = 2 and x = 4 is the only rational solution. All others are irrational.

Answer 2

Another way is to set x = y^a for real number a. To avoid trivial solutions a does not equal 1 or 0. Substitute this into

y^x = x^y to get y^(y^a) = (y^a)^y and equate exponents,

y^a = ay so that y^(a-1) = a and then y = a^{1/(a-1)} and

then x = y^a = a^{a/(a-1)} where as before, a is not 1 or 0.

These expressions are only integers for a=2 but they don't cover the solution x = y.

Answer 3

its pretty much impossible. its an absolute harsh b@@@@d of a question. consider for example the answer which a couple of folk have given already:
2^4=4^2; then consider 2^x=x^2. this has another root, which i know of no way to calculate exactly. by newton raphson, it is approximately -0.76
using maple, a computer algebra package solutions for x in terms of y are
x=- (y*LambertW(-ln(y)/y)) / ln(y)
where the LambertW function, well is a bit complicated. look it up on wikipedia, http://en.wikipedia.org/wiki/Lambertw
so there you go...

Answer 4

It relies upon on what I is the identification of, (XY, or YX), we can not assume that XY and YX are a similar (through fact matrix multiplication isn't communative) through fact that I is the identification matrix of one of the words, discover the inverse and multiply on the two sides then you definately can remedy for the matrix, or variable

Answer 5

y=x (x,y not equal to 0) or one is 2 and the other 4

ln(x)/x=ln(y)/y

Answer 6

I assume x and y must be different.

(x,y) = (2,4) or (4,2).

Answer 7

x ln(y) = y ln(x)<------take the ln of both sides

x = [yln(x)] / ln(y)
y= [xln(y)] / ln(x)

i hope that answeres it... using common sense, x=y