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 $(x_1,x_2)$

Question

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 $x_1x_2$.

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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?

Answer 1

HINT...We require both bracketed quadratics to be positive, which reduces to $2<x<3$. Combine this with the result you already have.

Answer 2

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 $5$