∵y=log₂x+log_x（2x）=log₂x+log_xx+log_x2
=log₂x+log_x2+1=log2x+

1

log2x

+1
令t=log₂x，∵x＞0且x≠1，∴t＞0或t＜0．
∴t+

1

t

≥2，或t+

1

t

≤−2
∴y=t+

1

t

+1≤-1，或y≥3，
故答案为：（-∞，-1]∪[3，+∞）．