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