设对数方程lg(ax)=2lg(x-1),讨论当a在什么范围取值时,该方程有解,并且求出它的解

lg(ax)=2lg(x-1),
=>ax=(x-1)²
且ax>0,x-1>0, =>a>0,x>1
=> x²-(a+2)x+1=0有解
=>Δ=(a+2)²-4≥0,
=> a≥0或a≤-4
=>a>0
解为 x=[a+2±√(a²+4a)]/2,
x>1
=>x=[a+2+√(a²+4a)]/2 , a>0