Question

Solve the following inequality:
$2^{{\log }_{2-x}(x^2+8x+15)}<1$

Answer 1

$2-x>02>xx\in (-\infty ,2)$ and $x^2+8x+15>0(x+5)(x+3)>0x\in (-\infty ,-5)\cup (-3,\infty )$

Taking intersection, we get $x\in (-\infty ,-5)\cup (-3,2)$

Taking ${\log }_2$ both side of equation

${\log }_{(2-x)}(x^2+8x+15)<0\frac{\log (x^2+8x+15)}{\log (2-x)}<0$

For $x\in (1,2)\log (2-x)<0$

$\log (x^2+8x+15)>0x^2+8x+15>1x^2+8x+14>0x\in (-\infty ,-4-\sqrt{2})\cup (-4+\sqrt{2},\infty )$

Hence, $x\in (1,2)$

For $x\in (-\infty ,-5)\cup (-3,1)\log (2-x)>0$

$\log (x^2+8x+15)<0x^2+8x+15<1x^2+8x+14<0x\in (-4-\sqrt{2},-4+\sqrt{2})$

Hence, $x\in (-4-\sqrt{2},-5)\cup (-3,-4+\sqrt{2})$

Taking union, we get $x\in (-4-\sqrt{2},-5)\cup (-3,-4+\sqrt{2})\cup (1,2)$