xy=yx
x=2, y=4.
The equation
xy=yx
x\ney,
(2,4)
(4,2),
(\tfrac{27}{8},\tfrac{9}{4})
(\tfrac{9}{4},\tfrac{27}{8})
y=vx.
J. van Hengel pointed out that if
r,n
r\geq3
rr+n>(r+n)r;
x=1
x=2
The problem was discussed in a number of publications. In 1960, the equation was among the questions on the William Lowell Putnam Competition,[1] which prompted Alvin Hausner to extend results to algebraic number fields.[2]
Main source:
An infinite set of trivial solutions in positive real numbers is given by
x=y.
aeb=c
a
b
a'ea'=c' ⇒ a'=W(c')
\begin{align} yx&=xy=\exp\left(ylnx\right)&\\ yx\exp\left(-ylnx\right)&=1&\left(multiplyby\exp\left(-ylnx\right)\right)\\ y\exp\left(-y
lnx | |
x |
\right)&=1&\left(raiseby1/x\right)\\ -y
lnx | \exp\left(-y | |
x |
lnx | |
x |
\right)&=
-lnx | |
x |
&\left(multiplyby
-lnx | |
x |
\right) \end{align}
⇒ -y
lnx | |
x |
=W\left(
-lnx | |
x |
\right)
⇒ y=
-x | |
lnx |
⋅ W\left(
-lnx | |
x |
\right)=\exp\left(-W\left(
-lnx | |
x |
\right)\right)
W(x)/x=\exp(-W(x))
Here we split the solution into the two branches of the Lambert W function and focus on each interval of interest, applying the identities:
\begin{align} W | ||||
|
\right)&=-lnx &for&0<x\lee,\\ W-1\left(
-lnx | |
x |
\right)&=-lnx &for&x\gee. \end{align}
0<x\le1
⇒
-lnx | |
x |
\ge0
\begin{align} ⇒ y&=
\exp\left(-W | ||||
|
\right)\right)\\ &=\exp\left(-(-lnx)\right)\\ &=x\end{align}
1<x<e
⇒
-1 | |
e |
<
-lnx | |
x |
<0
⇒ y=
\begin{cases} \exp\left(-W | ||||
|
\right)\right)=x\\ \exp\left(-W-1\left(
-lnx | |
x |
\right)\right) \end{cases}
x=e
⇒
-lnx | |
x |
=
-1 | |
e |
⇒ y=
\begin{cases} \exp\left(-W | ||||
|
\right)\right)=x\\ \exp\left(-W-1\left(
-lnx | |
x |
\right)\right)=x \end{cases}
x>e
⇒
-1 | |
e |
<
-lnx | |
x |
<0
⇒ y=
\begin{cases} \exp\left(-W | ||||
|
\right)\right)\\ \exp\left(-W-1\left(
-lnx | |
x |
\right)\right)=x \end{cases}
Hence the non-trivial solutions are:
Nontrivial solutions can be more easily found by assuming
x\ney
y=vx.
(vx)x=xvx=(xv)x.
\tfrac{1}{x}
x
v=xv-1.
The full solution thus is
(y=x)\cup\left(v1/(v-1),vv/(v-1)\right)forv>0,v ≠ 1.
Based on the above solution, the derivative
dy/dx
1
(x,y)
y=x,
(x,y)
(dy/dv)/(dx/dv),
dy | |
dx |
=
| ||||
v |
\right)
v>0
v ≠ 1.
Setting
v=2
v=\tfrac{1}{2}
42=24.
Other pairs consisting of algebraic numbers exist, such as
\sqrt3
3\sqrt3
\sqrt[3]4
4\sqrt[3]4
The parameterization above leads to a geometric property of this curve. It can be shown that
xy=yx
xv
v2
v ≠ 1
x8=y
82
(\sqrt[7]{8},\sqrt[7]{8}8),
xy=yx.
The trivial and non-trivial solutions intersect when
v=1
v=1
v\to1
v=1+1/n
n\toinfty
x=\limv\tov1/(v-1)=\limn\toinfty\left(1+
1n\right) | |
n |
=e.
y=x
xy-yx=0,y\nex
As
x\toinfty
y=1
y=1+
lnx | |
x |
+
3 | |
2 |
(lnx)2 | |
x2 |
+ … .
An infinite set of discrete real solutions with at least one of
x
y
x= | 1 |
\sqrt[3]{-2 |
y= | -2 |
\sqrt[3]{-2 |
-2
y=x
x<0
xx
x=y=-1
The equation
\sqrt[x]y=\sqrt[y]x
1/e
The curved section can be written explicitly as
| |||||||
y=e |
for 0<x<1/e,
W-1(ln(xx)) | |
y=e |
for 1/e<x<1.
This equation describes the isocline curve where power functions have slope 1, analogous to the geometric property of
xy=yx
The equation is equivalent to
yy=xx,
xy.
\sqrt[y]{y}=\sqrt[x]{x}
xy=yx
The equation
logx(y)=logy(x)