y=log2x+logx(2x) =log2x+log2(2x)/log2x =log2x+(1+log2x)/log2x =log2x+1/log2x+1 因为x>0且不等于1 所以log2x0 所以log2x+1/log2x≤-2或log2x+1/log2x≥2 y≤-2+1=-1或者y≥2+1=3 ∴函数的值域为(-∞,-1]∪[3,+∞)