Question

If the inequality ${\log }_a(x^2-x-2)>{\log }_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{9}{4}$ and if the solution set of the inequality is $(p,q)$, then value of $2(p+q)$ is

Answer 1

${\log }_a(x^2-x-2)>{\log }_a(-x^2+2x+3)$
For $\log$ to be defined
$x^2-x-2>0$
$\Rightarrow (x-2)(x+1)>0$
$\Rightarrow x\in (-\infty ,-1)\cup (2,\infty )\cdots \left(i\right)$
And
$-x^2+2x+3>0$
$x^2-2x-3<0$
$\Rightarrow (x+1)(x-3)<0$
$\Rightarrow x\in (-1,3)\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\Rightarrow x\in (2,3)\cdots \left(iii\right)$

Now, at $x=\frac{9}{4}$ given expression is satisfied. Substituting this in given inequality, we get
${\log }_a\left(\frac{13}{16}\right)>{\log }_a\left(\frac{39}{16}\right)$
which is possible when $0<a<1$
Given,
${\log }_a(x^2-x-2)>{\log }_a(-x^2+2x+3)$
$\Rightarrow x^2-x-2<-x^2+2x+3$
$\Rightarrow 2x^2-3x-5<0$
$\Rightarrow (x+1)(2x-5)<0$
$\Rightarrow x\in (-1,\frac{5}{2})\cdots \left(iv\right)$

Therefore, from $\left(iii\right)$ and $\left(iv\right)$ we get,
$x\in (2,\frac{5}{2})$
$\because (p,q)\equiv (2,\frac{5}{2})$
$\therefore 2(p+q)=2(2+\frac{5}{2})=9$