Are there real solutions to $x^y = y^x = 3$ where $y \neq x$?

Question

I need to solve the following equation for (x,y) $$x^y = y^x = 3$$

Everytime I run a numerical method for this problem, I get $$ (x,y) = (1.82546...,1.82546..) $$

I expect there to be a solution where $x \neq y$ but I cannot seem to find it by any means.

Does such a solution exist?

Answer 1

Assume $0<x<y$ and $x^y=y^x=3$. Then in particular $\frac{\ln x}x=\frac{\ln y}y$. The derivative of $f(t)=\frac{\ln t}t$ is $f'(t)=\frac{\frac1t\cdot t-1\cdot \ln t}{t^2}=\frac{1-\ln t}{t^2}>0$ for $0<t<e$, hence $f$ is injective on $(0,e]$. We conclude that $y>e$. Also note that for $x,y>0$ we have $x^y\le 1$ if $x\le 1$. Therefore $x>1$.

From $x>1$ and $y^x=3$ it follows that $y<3$. Then $$x=\sqrt[3]{x^3}\ge \sqrt[3]{x^y}=\sqrt[3]{y^x}>\sqrt[3]y>\sqrt[3]e>1+\frac13+\frac1{18}=\frac{25}{18}>\frac54.$$ Thus $$y^x>e^{\frac54}>1+\frac54+\frac{25}{32}=\frac{97}{32}>3,$$ contradiction.

---

Remark: The above does not require any calculator. All "numerical" calculations have been reduced to adding and multiplying fractions with manageably small denominator, and we use that for $t>0$ we have $e^t=\sum_{k=0}^\infty\frac{t^k}{k!}>1+t+\frac{t^2}2$.

Of course, if one trusts in calculator results, $\sqrt[3]e\approx 1.395612425$ and $e^{\sqrt[3]e}\approx4.037446449$ and that is of course also $>3$.

Answer 2

Here is an interesting way that works for finding the positive real solutions to $x^y = y^x = c$ for any real $c \in (1,e^e)$. $\def\rr{\mathbb{R}}$

Take any reals $x,y$.

Firstly, $y^x = c$ implies $y = c^{1/x}$, and since $( t \mapsto c^{1/t} )$ is a strictly decreasing function on $\rr^+$ with range $\rr^+$, it has at least one fixed point $z$, namely $z \in \rr^+$ such that $z = c^{1/z}$. Now clearly $(z,z)$ is a solution.

Secondly, $x^y = y^x = c$ implies $x^{c^{1/x}} = c$, but $( x \mapsto x^{c^{1/x}} )$ is a strictly increasing function on $\rr^+$, and so there is at most one solution for $x$ and hence at most one solution for $(x,y)$.

Therefore we have found the only solution, which must be $(z,z)$.

Technical detail

To prove the strictly increasing nature of the function in the second paragraph above, note that $\frac{d}{dx}( x^{c^{1/x}} ) = \frac{d}{dx}( e^{\ln(x)c^{1/x}} )$ $= e^{\ln(x)c^{1/x}}( \frac{1}{x} c^{1/x} - \ln(x) c^{1/x} \ln(c) \frac{1}{x^2} )$ $= e^{\ln(x)c^{1/x}} c^{1/x} \frac{1}{x^2} ( x - \ln(x) \ln(c) )$ $\ge 0$ because $x - \ln(x) \ln(c) > x - \ln(x) e \ge 0$ which can be proven by noting that $\frac{d}{dx}( x - \ln(x) e ) = 0$ iff $1 - \frac{e}{x} = 0$ iff $x = e$, which is a global minimum because $\frac{d}{dx^2}( x - \ln(x) e ) = \frac{e}{x^2} > 0$.

Answer 3

The solution may only be obtained using the Lambert W function for other solutions, which must be complex.

$$x^y=y^x\implies y=e^{W_k(x\ln(x))}$$

Trivially, there is $y=e^{W_0(x\ln(x))}=x$

But there are other branches, as you can see graphically.

I will go about trying to solve:

$$y^x=3$$

$$(e^{W_k(x\ln(x))})^x=e^{xW_k(x\ln(x))}=3$$

$$xW_k(x\ln(x))=3$$

$$W_k(x\ln(x))=\frac3x$$

$$x\ln(x)=\frac3xe^{\frac3x}$$

$$x^x=e^{\frac3xe^{\frac3x}}$$

We have the simple case $x=e^{\frac3x}$, or we don't have that case.

Either way, it is impossible to solve it from here in closed form using the Lambert W function.

I would proceed by attempting to solve the last equality for complex $x$, as it is obvious there are no other real solutions.

Answer 4

I will show that the smallest value of $z$ such that there are an $x$ and $y$ with $x^y = y^x = z$ is $z = e^e \approx 15.15426224 $, so there is no solution to OP's question.

This is a corrected version, which should ameliorate user21820's righteous indignation.

Start with the usual parameterization of $x^y = y^x$: Let $y = rx$ where $r > 1$. Note that this implies $x < e < y$.

Then $x^{rx} = (rx)^x$ or $x^r = rx$ or $x^{r-1} = r$ or $x = r^{1/(r-1)}$ and $y = rx =r^{1+1/(r-1)} =r^{r/(r-1)} $.

Then $x^y =(r^{1/(r-1)})^{r^{r/(r-1)}} =r^{r^{r/(r-1)}/(r-1)} $.

As a check, $y^x =(r^{r/(r-1)})^{ r^{1/(r-1)}} =r^{r^{1/(r-1)}r/(r-1)} =r^{r^{r/(r-1)}/(r-1)} $.

If $r = 1+s$, this is

$\begin{array}\\ f(s) &=(1+s)^{(1+s)^{(1+s)/s}/s}\\ &=(1+s)^{(1+s)^{1+1/s}/s}\\ &=(1+s)^{e^{\ln(1+s)(1+1/s)}/s}\\ \text{so}\\ g(s) &=\ln(f(s))\\ &=\ln(1+s)e^{\ln(1+s)(1+1/s)}/s\\ &=\frac{\ln(1+s)}{s}e^{\ln(1+s)(1+1/s)}\\ &=\frac{\ln(1+s)}{s}e^{\ln(1+s)+\ln(1+s)/s}\\ &=(1+s)\frac{\ln(1+s)}{s}e^{\ln(1+s)/s}\\ &=e+(e s^2)/24-(e s^3)/24+(73 e s^4)/1920+O(s^5)\\ &\qquad\text{according to Wolfy}\\ &=e\left(1+( s^2)/24-( s^3)/24+(73 s^4)/1920+O(s^5)\right)\\ \text{so}\\ f(s) &=e^e\left(1+s^2/24-s^3/24+(7 s^4)/180+O(s^5)\right)\\ &\qquad\text{again, according to Wolfy}\\ \end{array} $

I will now show that $f(s)$ is increasing by showing that $g(s)$ is increasing.

Using Wolfy again, $g'(s) = \dfrac{((s+1)^{1/s} (s^2-(s+1) \ln^2(s+1)))}{s^3} $.

To show that $g'(s) > 0$ for $s > 0$, we need to show that $h(s) =s^2-(s+1) \ln^2(s+1) \gt 0 $ for $s > 0$. $h(0) = 0$ and $h'(s) = 2 s-\ln^2(s+1)-2 \ln(s+1) $.

We need to do the same with $h'(s)$. $h'(0) = 0$ and, for $s > 0$, $h''(s) = 2 \frac{(s-\ln(s+1))}{s+1} \gt 0 $ since $\ln(1+s) < s$.

Therefore $f'(s) > 0$ for $s > 0$. Since $f(0) = e^e$, the smallest value for which $x^y = y^x$ is $e^e \approx 15.15426224 $, so this can not be $3$.

This is why $2^4 = 4^2 =16$ works.