Logarithmic inequality

log_x(x^3-2x) > 3

I'm crying right now :P What should I do with this?

I've come as far as to state that valid values for x are > sqrt(2), but when I try to solve log_x(x^3-2x) > 3 I end up with x being < 0 :/

x^3 - 2x > x^3
right...?

(x>0)AND(x=/=1)AND(x^3-2x)>0

When this is fulfilled:

x^3-2x>x^3
-2x>0
x<0

Then x belongs to an empty set.