The derivative f'(x)= 2ln(x-2)+2x/(x-2),

When 4 >; X> at three o'clock, f'(x)>0.

Therefore, f(x) is the increasing function in the interval (3,4).

f(3)=-3 & lt； 0,

f(4)=8*ln2-3=ln(256/e？ )> is a natural alkali,

Therefore, f(x) must have a zero point in the interval (3,4).