x^y = y^x

Trivial Solutions:


Based on some algebraic considerations solutions can be found:

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

assumption:y=kx assumption: y= kx

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

(x)(kxx)=(kx)(xx) (x)^{(\frac{kx}{x})}= (kx)^ {(\frac{x}{x})}

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

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

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

(x)(k1)=(k) (x)^{(k-1)}= (k)

x=(k)(1k1) x= (k)^{(\frac{1}{k-1})}

x=kk1x= \sqrt[k-1]{k}

y=kkk1y= k \sqrt[k-1]{k}

(kk1)(kkk1)=(kkk1)(kk1) (\sqrt[k-1]{k})^{(k\sqrt[k-1]{k})}= (k\sqrt[k-1]{k})^ {(\sqrt[k-1]{k})}


When we let k=2k=2 we get y=4y=4 and x=2x =2. 24=422^{4} = 4^{2} 16=1616=16

When we let k=12k =\frac{1}{2} we get the inverse relationship where x=4x=4 and y=2y=2.


When we let k=4 we get x=43x =\sqrt[3]{4} and y=443y=4\sqrt[3]{4}

(43)(443)=(443)(43) (\sqrt[3]{4})^{(4\sqrt[3]{4})}= (4\sqrt[3]{4})^ {(\sqrt[3]{4})} 18.805... = 18.805...

When we let k=14k=\frac{1}{4} we get the inverse scenario where x=443x=4\sqrt[3]{4} andy=43y =\sqrt[3]{4}

There are some restriction on kk. The functions for xx and yy are undefined when k=0k=0 since you it results in x=011=01=10x=0^{\frac{1}{-1}}=0^{-1}=\frac{1}{0} and y=0011=010y=0*0^{\frac{1}{-1}}=0*\frac{1}{0} and since we can't divide by zero this is undefined. Interestingly though 000^{0} is defined as 00=10^{0} =1 and therefore 00=000^{0}=0^{0}. Although this is outside our solution set since no value of kk seems to produce this result. Other cases that are undefined include the case where k=1k=1 since it again results in a division by zero. The solution 11=111^{1}=1^{1} is also outside this solution set since there is no value of kk that produces one for xx and one for yy. 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=1k=-1 we get y=1111=112=undefinedy= -1 \sqrt[-1-1]{-1}= -1 \sqrt[-2]{-1}= undefined and x=111=12=undefinedx= \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. yy is in blue and xx is in green. As we can see xxand yy have the same domain for kk but they have different ranges. The one sided limit for yy 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 kk?

Ruilin Wang - 3 weeks, 5 days ago

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Just as another visualization aid:

If x,y>0,x,y>0, you can take the log of both sides, and get lnxx=lnyy. \frac{\ln x}{x} = \frac{\ln y}{y}. Drawing horizontal lines on the graph of lnxx\frac{\ln x}{x} shows that for any positive value of z1/e,z \le 1/e, there are exactly two values x,yx,y with lnxx=lnyy=z.\frac{\ln x}{x} = \frac{\ln y}{y} = z.

Patrick Corn - 3 weeks, 2 days ago

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

Ruilin Wang - 3 weeks, 5 days ago

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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.

Brody Acquilano - 3 weeks, 3 days ago

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

Brody Acquilano - 3 weeks, 3 days ago

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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.

Brody Acquilano - 2 weeks, 6 days ago

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“0^0 is defined as 1” Respect!

Kano Boom - 2 weeks, 4 days ago

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