Another one of those inequalities

Well, first of all, $3-2x > 0 \\ 4x+1>0 \\ \therefore x \in \left(-\frac{1}{4},\frac{2}{3}\right)$ Now, $\log_5(3-2x) \ge \log_5(4x-1) \\ 3-2x \ge 4x-1 \\ 6x \le 2 \\ x \le \frac{1}{3}$ Combining these inequalities, we get $x \in (-\frac{1}{4},\frac{1}{3}] \\ a=\frac{1}{4} \text{ and } b=\frac{1}{3} \\ a \times b= \frac{1}{12} = \boxed{0.083}$