Functions question - find x

Find {x:x^(3/2) > x^2} and {x:x^(-3/2) , x^(-2)}.

Is there a basic formula or something that provides the answer?

first one
x^(3/2) > x^2
[X^(3/2)]/(x^2) > 1 <-since x^2 is always positive
x^(-1/2) > 1
1/sqrt(x) > 1
sqrt(x) < 1
0 < x < 1

there is nothing to find in the second one, but if you meant < instead of ,
x^(-3/2) < x^(-2)
1/[x^(3/2) * x^(-2)] < 1 <- since x^(-2) is always positive
1/[x^(-1/2) < 1
sqrt(x) < 1
0 < x < 1

just go lyk dis
x^(3/2)>x^2
x^(3/2)-x^2>0
x^(3/2)(1-x^1/2)>0

your second eqn is nt clear...