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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?

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    $\begingroup$ What makes you expect that there is another solution? $\endgroup$ – Henning Makholm Feb 20 '16 at 15:23
  • $\begingroup$ I run $x^y=y^x$ numerically and draw it, it seems the program gives $x=y$ when $0<x<e$. I think it does not exist. $\endgroup$ – Alexis Feb 20 '16 at 15:36
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    $\begingroup$ @SS_C4: Graphing is not a valid reason. $\endgroup$ – user21820 Feb 20 '16 at 15:56
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    $\begingroup$ But it does give the answer. $\endgroup$ – SS_C4 Feb 20 '16 at 15:56
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    $\begingroup$ @SS_C4: No it does NOT. If not I can also say that the graphs of $y = 2$ and $y = x^2$ do not have rational intersections. It's obvious from looking at the graphs!!! $\endgroup$ – user21820 Feb 20 '16 at 16:05
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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 $>3$.

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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.

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  • $\begingroup$ Did you not read both my answer and Hagen's? It's totally invalid to use asymptotic approximations to solve an equation anywhere except at the point of approximation, so your post is logically in error. For that reason I didn't check any of your approximations, and besides you also acknowledge that the approximation is not valid for the $x,y$ that we are interested in. If you want, you'd have to use concrete inequalities, not approximations. $\endgroup$ – user21820 Feb 21 '16 at 8:27
  • $\begingroup$ I have it right now. $\endgroup$ – marty cohen Feb 21 '16 at 18:11
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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 $\frac{d}{dx^2}( x - \ln(x) e ) = \frac{e}{x^2} > 0$.

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  • $\begingroup$ Doy you mean that for $c=16$ there is also only one solution, although $2^4=4^2=16$? $\endgroup$ – Hagen von Eitzen Feb 20 '16 at 16:34
  • $\begingroup$ @HagenvonEitzen: Sorry it works only for $c \in (1,e^e)$ and I've fixed my answer and added the proof. $\endgroup$ – user21820 Feb 20 '16 at 16:49
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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.

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