Question

x^y=y^x and x=2y then find the value of x and y

Answer 1

Given- X^y=y^x and x=2y

we can write-
    x^y/ y^x=1

now put x=2y then

(2y)^(y)/ y^(2y) =1

=> (2y)^y = y^(2y) 
indivisually separate  the terms..
=>(2^y) *(y^y) = (y^2)^y
=>(2^y) *(y^y) = (y*y)^y
=>(2^y) *(y^y) = (y^y)*(y^y)
=> (2^y) *(y^y) - (y^y)*(y^y) =0
=>(y^y)[(2^y)- (y^y) ]=0 
now from the following equations we get 
either y^y=0 or (2^y)- (y^y)=0
Sincey is not zero, y^y is not zero.
now, (2^y)- (y^y)=0
 2^y =y^y 
comparing LHS and RHS we get,
y = 2 
x = 2y = 4
x=4   Answer