Name: Anonymous 2007-11-29 15:17
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...?
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...?