7. Let $0<u \neq 1$, and $f(x)=\log _{a}\left(6 a x^{2}-2 x+3\right)$ is monotonically increasing on $\left[\frac{3}{2}, 2\right]$, then the range of values for $a$ is $\qquad$ .
7. $a \in\left(\frac{1}{24}, \frac{1}{12}\right] \cup(1,+\infty)$

From $6 a x^{2}-2 x+3>0 \Leftrightarrow a>-\frac{1}{2 x^{2}}+\frac{1}{3}=-\frac{1}{2}\left(\frac{1}{x}-\frac{1}{3}\right)^{2}+\frac{1}{18}$. Since $\frac{3}{2} \leqslant x \leqslant 2$, i.e., $\frac{1}{2} \leqslant \frac{1}{x} \leqslant \frac{2}{3}$. Thus $a>\frac{1}{24}$.
(1) When $\frac{1}{24}<a \leq \frac{1}{12}$, $f(x)$ is monotonically increasing on $\left[\frac{3}{2}, 2\right] \Rightarrow g(x)=6 a x^{2}-2 x+3$ is monotonically increasing on $\left[\frac{3}{2}, 2\right]$. The increasing interval of $g(x)$ is $\left[\frac{1}{6 a},+\infty\right)$. Therefore, $\frac{1}{6 a} \leqslant \frac{3}{2}, a \geq \frac{1}{9}$, hence $a>1$.
So $a \in\left(\frac{1}{24}, \frac{1}{12}\right] \cup(1,+\infty)$.