7. Given the set $A=\left\{x \mid \log _{a}(a x-1)>1\right\}$. If $2 \in A$, then the range of values for $a$ is $\qquad$
II.7. $\left(\frac{1}{2}, 1\right) \cup(1,+\infty)$.
From the problem, we get $\log _{a}(2 a-1)>1$.
Then $\left\{\begin{array}{l}0<a<1, \\ 2 a-1>1, \\ 2 a-1>a \text {. }\end{array}\right.$
Solving, we get $\frac{1}{2}<a<1$ or $a>1$.