If the inequality $\log_a(x^2-x-2)>\log_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{9}{4}$ in the interval $(x_1,x_2)$ If the inequality $\log_a(x^2-x-2)>\log_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{9}{4}$ in the interval $(x_1,x_2)$,then find the product $x_1x_2$.

Here $a$ is not specified .I know that $\log$ is an increasing function,so we can cancel it out from both the sides,
$\log_a(x^2-x-2)>\log_a(-x^2+2x+3)$
$(x^2-x-2)>(-x^2+2x+3)$
$2x^2-3x-5>0$
$(x+1)(2x-5)>0$
So $x\in(-\infty,-1)\cup(\frac{5}{2},\infty)$ but $\frac{9}{4}=2.25$ is not in this interval.
What wrong have i done in this?What is the correct way to solve this and what is the correct answer?

You have done interval for $x<0$ plug in $-\infty$ you will see why. So the correct interval is $(2,5/2)$ so the product is $5$
