Is there a closed form?

Calculus Level 5

It is given that there is a value $x$ such that

${x}^{y} = {y}^{x}\ \ (x>1)$

has only $1$ solution for $y$ . Let that solution for $y$ be $A$ .

Find $\left\lfloor 10000A \right\rfloor.$

Details and Assumptions

If you think that there is more than $1$ such value for $y$ , input -1.
If you think that there is no such value for $x$ , input 2.
$x$ and $y$ are real.

The answer is 27182.

1 solution

Julian Poon
Jan 13, 2015

For this question, I can't find a closed form for $y$ . However, it is still solvable.

Due to symmetry, there would always be the solution $x=y$ . However, that is not the only solution for mast numbers.

The expression can be written as $\frac { \ln{(x)} }{ x } =\frac {\ ln{(y)} }{ y }$

Now let $\frac { \ln{(y)} }{ y } =a$ , so $\frac { \ln{(x)} }{ x } = a$

You can see from this that when $x<1$ , there is only $1$ solution which is $x=y$ , since it means that $a$ is a negative and that $a$ would not be negative even as $y$ approaches infinity.

Now, for $1$ solution to occur other than when $x<1$ , we would have to find the maximum for $a$ . So,

$\frac { d }{ dx } \frac { \ln{(x)} }{ x } =\frac { 1-\ln{(x)} }{ { x }^{ 2 } } =0$ , $x = e$

Now due to symmetry, there would always be the soultion $x=y$ , so $y=e$

Therefore, $\left\lfloor 10000A \right\rfloor =\left\lfloor 10000e \right\rfloor = \boxed{27182}$

Here is a graphic of what I am talking about: here

General parametric solution

$\displaystyle x = \left( 1 + \frac{1}{k} \right)^{k}$

$\displaystyle y = \left( 1 + \frac{1}{k} \right)^{k+1}$ .

Krishna Sharma - 6 years, 5 months ago

Wow. How did you derive that?

Julian Poon - 6 years, 5 months ago

First substitute $y = mx$ to obtain

$\displaystyle x = m^{\frac{1}{m-1}}$

$\displaystyle y = m^{\frac{m}{m-1}}$

These are also parametric solutions, further substitute $k = \dfrac{1}{m-1}$ to obtain the above result :)

Krishna Sharma - 6 years, 5 months ago

This is a nice problem! I believe there should be a proof by the Lambert W-function, but I went with my intuition and followed a proof a bit like yours to get $x = y = e$ .

Jake Lai - 6 years, 5 months ago

Yeah. But Lambert W-function seems beyond me.

Julian Poon - 6 years, 5 months ago

what about $4^{2}$ = $2^{4}$ ? Surely that makes for another solution?

Curtis Clement - 6 years, 5 months ago

When $x=2$ , $y$ can be $2$ or $4$ or $-0.7666646959...$ So, $x=2$ isnt a solution because there is more than $1$ solution for $y$ .

However, $y=x=e$ is the answer because when $x=e$ , there is no other solutions for $y$ .

Julian Poon - 6 years, 4 months ago

@Julian Poon $\textbf{PLEASE HELP!}$ This is how I had tried at first. ${x}^{y} = {y}^{x}$ ${x}^ {\frac{y}{x}} = y$ ${y}^ {\frac{x}{y}} = x$ $Taking \frac{dy}{dx} and \frac{dx}{dy}$

$\frac{dy}{dx} = \frac{d}{dx}\ {{x}^ {\frac{y}{x}} = {\frac{y}{x}} }{x}^{\frac{x}{y} - {1}} = {yx}^ {\frac{x-y}{y}}$ $\frac{dx}{dy} = \frac{d}{dy}\ {{y}^ {\frac{x}{y}} = {\frac{x}{y}} }{y}^{\frac{y}{x} - {1}} = {xy}^ {\frac{y-x}{x}}$ Using $\frac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}}$ and plugging in the values of $\frac{dy}{dx}$ and $\frac{dx}{dy}$ , we get

${yx}^ {\frac{x-y}{y}} = \dfrac{1}{{xy}^ {\frac{y-x}{x}}}$ ${yx}^ {\frac{x-y}{y}}\times {xy}^ {\frac{y-x}{x}} = 1$

${x}^ {\frac{y}{x}} {y}^ {\frac{x}{y}} = 1$ Replacing with x and y, $yx=1$

Now, for $x>1$ , $y$ can assume infinite fractional values as solutions.

Could somebody please explain why what I've done is wrong? Thanks very much in advance.

User 123 - 6 years ago

Your differentiation is incorrect, because the other variable is not a constant.

Kenny Lau - 5 years, 8 months ago

y = x = 2 is a solution TOO !!!

Ossama Ismail - 6 years, 5 months ago

Yes but when $x=2$ , there is another solution which is $y = 4$ and $y=-0.7666646959...$

Julian Poon - 6 years, 5 months ago

Is there a closed form?

The answer is 27182.

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