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.

earl w. swokowski has the answer

>>2

That was also my approach. My supervisor keeps telling me I'm doing it wrong though :/

>>1
log_x(x^3-2x)>3
x^3-2x>x^3
-2x>0
x<0
>>4 if your professor said you did it wrong, he's wrong.

>>2

He still won't budge :D

The function works only for x > sqrt(2).
And f(x)=3 is its asymptote, so it's never > 3.