已知函数f(x)=lg(ax-1)-lg(x-1)在区间[10,正无穷）上单调递增,求实数a的取值范围

f(x)=lg(ax-1)-lg(x-1)在区间[10,正无穷）f(x)=lg(ax-1)-lg(x-1) =lg [ (ax-1)/ (x-1) ]=lg [[ (a(x-1) +(a -1) ]/(x-1)]=lg ( a + (a-1)/(x-1)]令：y = lgu,u =a + (a-1)/ (x-1)u =a + (a-1)/ (x-1) 在【10,正无穷...请问u=a+(a-1)/9为什么是大于0的u=a+(a-1)/9它是对数的真数，真数就应该大于0,否则对数没有意义