3A. Solve the inequality $\log _{x^{2}}(3-2 x)>1$.
Solution. First, we note that $x \neq 0$ and $3-2 x>0$. The inequality is equivalent to $\log _{x^{2}}(3-2 x)>\log _{x^{2}} x^{2}$. We will consider two cases:

1) $0<x^{2}<1$ i.e. $x \in(-1,0) \cup(0,1)$

The inequality reduces to the system

$$
\left\{\begin{array}{l}
3-2 x>0 \\
x^{2}>3-2 x
\end{array}\right.
$$

which has no solution on the above interval.
2) $x^{2}>1$ i.e. $x \in(-\infty,-1) \cup(1,+\infty)$

The inequality reduces to the system

$$
\left\{\begin{array}{l}
3-2 x>0 \\
x^{2}<3-2 x
\end{array}\right.
$$

which, taking into account the above interval, has the solution interval ( $-3,-1$ ). Finally, the solution to the original inequality is the interval ( $-3,-1$ ).