Solving x & y Equation: Find x for y>x in R

[vishal007win]

if x^y=y^x,
i) x and y are integers such that y > x, find the solutions?
ii) for y>x , x, y belong to R , find the range of x for which this equation has real solutions?


for first part i did,
taking log on both side
ln y/ln x = y/x
so if y and x are integers, term on right is a rational number,
let t=y/x >1
implies ln y/ln x= t
so y = x^t
which on putting back we get,
t= x^t-1
so for x an integer and t a rational number greater than 1, by hit and trial
one can say
first case is
x=2 and t=2
so y =4

no other possible solution
now how to proceed for 2nd part??

 

[AC130Nav]

Not doing the math, it seems you already solved ii. The first problem (i) does not specify y>x, so all x=y are part of the solution set.

 

[vishal007win]

in part i) it is specified .. that y>x ,
unable to solve 2nd part..hope to get some hint from here..

 

[Wizlem]

Are you restricting x and y to positive values? This problem would appear to become quite tricky otherwise. For instance -2 and -4 are solutions also. If you can assume x>0 consider the function f(z) = z^(1/z). It's not hard to show this function is continuous and increases to a point and then decreases. Then f(z)=a for some a only has at most 2 solutions for z. If x<y, x must be less than the point where f(z) is maximum.

 

[vishal007win]

yes i m restricting only to positive cases.
i thought that way too,
its like ,
let a f(x)=x^(1/x), function increases till x=e , and then goes down, and is never negative for positive values of x,
so now if
x₁ ^(1/x₁)=x₂ ^(1/x₂) ---1

where x₁<x₂, will solve my problem
then it gives me solution that for all values of x₁<e , there will exist an x₂ satisfying relation (1)

is this solution correct??

 

[Wizlem]

f(x)=x^(1/x)>1 for all x>e i believe so you have to restrict yourself to x such that f(x)>1.