f(x)=lg[(ax-1)/(x-1)]
看真数即可
令u(x)=(ax-1)/(x-1)=[a(x-1)+a-1]/(x-1)=a+(a-1)/(x-1)
要在[10,+∞)上递增,则a-10,得：a>1/10
所以,实数a的取值范围是(1/10,1)