Solve ${(x^y=y^x),(x^2=y^3):}$ ?

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Answer 1 · 2016-12-10T12:47:09

$((x=0, y=0), (x=1,y=1), (x=27/8, y = 9/4))$

$x^y=y^x$ admits as solution $x=y$

Now applying $log$ to the equations

${(x^y=y^x),(x^2=y^3):}$ -----(1)

we obtain

${(y log x=x log y),(2 logx=3 logy):}$

Dividing term to term we obtain the relationship

$y/2=x/3$

The solutions for (1) are obtained solving

${(x=y),(x^2=y^3):}$

and

${(x^2=y^3),(y/2=x/3):}$

so they are

$((x=0, y=0), (x=1,y=1), (x=27/8, y = 9/4))$

Attached a figure showing the interceptions

In red the two leafs of $x^y=y^x$
blue dotted $x=y$
blue continuous $y/2=x/3$ and
black $x^2=y^3$

The intersections have a black dot over.

The solution at $0,0$ can be understood as a regularization.

enter image source here

Solve {(x^y=y^x),(x^2=y^3):}{(x^y=y^x),(x^2=y^3):} ?