Question

Solving the inequation
${\log }_{\left(\frac{25-x^2}{16}\right)}\left(\frac{24-2x-x^2}{14}\right)>1$ we get $x\in \left(\right(k,y)\cup (z,w\left)\right)$.Find $k+y+z+w$?

Answer 1

The given inequaiton is valid for
$\frac{24-2x-x^2}{14}>0and\frac{25-x^2}{16}>0\therefore x^2+2x-24<0and{x}^2-25<0\Rightarrow (x+6)(x-4)<0and-5<x<5\Rightarrow -6<x<4and-5<x<5$
combining both inequality
$-5<x<4......\left(1\right)$
Now consider the following cases:
Case I: If $0<\frac{25-x^2}{16}<1\Rightarrow 9<x^2<25$
$\therefore xϵ(-5,-3)\cup (3,5).........\left(2\right)$
$\therefore$ The given inequation convert in the form
$\frac{24-2x-x^2}{14}<\frac{25-x^2}{16}$
$\Rightarrow x^2+16x-17>0$
$\therefore xϵ(-\infty ,-17)\cup (1,\infty ).........\left(3\right)$
combining (2) and (3), we get
$\therefore xϵ(3,5)but-5<x<4$ {from(1)}
$\therefore xϵ(3,4)$ ...........(4)
Case II: If $\frac{25-x^2}{16}<1\Rightarrow xϵ(-3,3)......\left(5\right)$
$\therefore$ The given inequation convert in the form
$\frac{24-2x-x^2}{14}>\frac{25-x^2}{16}$
$\frac{24-2x-x^2}{14}>\frac{25-x^2}{16}<1\Rightarrow xϵ(-3,3)......\left(5\right)\Rightarrow x^2+16x-17<0\therefore xϵ(-17,1)...........\left(6\right)$
combining (5) and (6), we get
$\therefore xϵ(-3,1)but-5<x<4\left\{from\left(1\right)\right\}\therefore xϵ(-3,1)$
Now combining (4) and (7), the inequaiton have the final solution is
$xϵ(-3,1)\cup (3,4)$