# x^y = y^x

Trivial Solutions:

$1^{1}=1^{1}$

Based on some algebraic considerations solutions can be found:

$(x)^{(y)}= (y)^ {(x)}$

$assumption: y= kx$

$(x)^{(kx)}= (kx)^ {(x)}$

$(x)^{(\frac{kx}{x})}= (kx)^ {(\frac{x}{x})}$

$(x)^{(k)}= (kx)^ {(1)}$

$(x)^{(k)}= (kx)$

$(\frac{x^{(k)}}{x})= (\frac{kx}{x})$

$(x)^{(k-1)}= (k)$

$x= (k)^{(\frac{1}{k-1})}$

$x= \sqrt[k-1]{k}$

$y= k \sqrt[k-1]{k}$

$(\sqrt[k-1]{k})^{(k\sqrt[k-1]{k})}= (k\sqrt[k-1]{k})^ {(\sqrt[k-1]{k})}$

ex1.

When we let $k=2$ we get $y=4$ and $x =2$. $2^{4} = 4^{2}$ $16=16$

When we let $k =\frac{1}{2}$ we get the inverse relationship where $x=4$ and $y=2$.

ex2.

When we let k=4 we get $x =\sqrt[3]{4}$ and $y=4\sqrt[3]{4}$

$(\sqrt[3]{4})^{(4\sqrt[3]{4})}= (4\sqrt[3]{4})^ {(\sqrt[3]{4})}$ 18.805... = 18.805...

When we let $k=\frac{1}{4}$ we get the inverse scenario where $x=4\sqrt[3]{4}$ and$y =\sqrt[3]{4}$

There are some restriction on $k$. The functions for $x$ and $y$ are undefined when $k=0$ since you it results in $x=0^{\frac{1}{-1}}=0^{-1}=\frac{1}{0}$ and $y=0*0^{\frac{1}{-1}}=0*\frac{1}{0}$ and since we can't divide by zero this is undefined. Interestingly though $0^{0}$ is defined as $0^{0} =1$ and therefore $0^{0}=0^{0}$. Although this is outside our solution set since no value of $k$ seems to produce this result. Other cases that are undefined include the case where $k=1$ since it again results in a division by zero. The solution $1^{1}=1^{1}$ is also outside this solution set since there is no value of $k$ that produces one for $x$ and one for $y$. You will find that k can also not be a negative number since it results in negative roots and that is an operation that is undefined.

ex 3.

When we let $k=-1$ we get $y= -1 \sqrt[-1-1]{-1}= -1 \sqrt[-2]{-1}= undefined$ and $x= \sqrt[-1-1]{-1}= \sqrt[-2]{-1}= undefined$

For a similar reason k can not be a complex number. Complex numbers also produce undefined results using this method. Taking a complex root is undefined.

If we look at a graph of the equation for x and y we can see this clearer. $y$ is in blue and $x$ is in green. As we can see $x$and $y$ have the same domain for $k$ but they have different ranges. The one sided limit for $y$ as k approaches zero appears to be one. That means as k gets smaller and smaller y will get closer to one. On the other hand if we consider what happens to x it seems to behave in the opposite manner. It apporaches infinity as k approaches zero. This explains the behavior we saw earlier with the inverses of k. As k changes x and y have to swap places. If y starts out larger than x and we decrease y eventually y will equal x. We find at k=1.643 x=y =3.556. As y decreases x will increase to maintain the relationship. Unfortunately the fact that x and y do not have the same range means that y can not be less than one but x can which is not exactly true because we know we can just flip x and y and the equation will still look the same. So we know simply by looking at the graph of x and y that this does not include the full solution set.

Note by Brody Acquilano
3 weeks, 5 days ago

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Is there any restriction for $k$?

- 3 weeks, 5 days ago

Just as another visualization aid:

If $x,y>0,$ you can take the log of both sides, and get $\frac{\ln x}{x} = \frac{\ln y}{y}.$ Drawing horizontal lines on the graph of $\frac{\ln x}{x}$ shows that for any positive value of $z \le 1/e,$ there are exactly two values $x,y$ with $\frac{\ln x}{x} = \frac{\ln y}{y} = z.$

- 3 weeks, 2 days ago

Wow, I had been wondering this for a long time. Thanks!

- 3 weeks, 5 days ago

I have updated the note and will continue to update it as my knowledge on this subject improves. I have included some of the restriction on k. Although I by no means speak authoritatively on this subject. I also have been wondering this for a long time and only recently stumbled upon this solution.

- 3 weeks, 3 days ago

If anyone has a better solution or more complete solution please indulge me.

- 3 weeks, 3 days ago

I concur. I noticed a similar result with the intersections of exponential growth functions and polynomial functions. i.e. x^2 = 2^x. One function is always exponential,either growth or decay, and the other should be a polynomial ,of even or odd degree (which affects end behavior). I didn't exactly try every combination but for the ones I tried I noticed two intersections.

- 2 weeks, 6 days ago

“0^0 is defined as 1” Respect!

- 2 weeks, 4 days ago