Find the number of values of x(x is real) for which 64^1/x+48^1/x=80^1/x

Given, 64^1/x+48^1/x = 80^1/x

Let y= 1/x

Then the equation is 64^y+48^y = 80^y

Divide by 16^y on both sides, we get

3^y + 4^y = 5^y

It is of form

sin^yz + cos^yz =1,    where sinz=3/5,cosz=4/5

now above equality is true iff y=2, Since sinz≠0,±1

→ x=1/2

hence number of solutions is 1.