Question

Solve the following inequality:
${\log }_{1/4}(x+1)⩾-2{\log }_{1/16}2+{\log }_{1/4}(x^2+3x+8)$

Answer 1

${\log }_{1/4}(x+1)\ge -2{\log }_{1/16}\left(2\right)+2{\log }_{1/4}(x^2+3x+8)$

Here $x+1>0$ & $(x^2+3x+8)>0$

$x>-1xϵR$

Hence from here $xϵ(-1,\infty )$

${\log }_{1/4}(x+1)\ge -2{\log }_{1/16}\left(2\right)+{\log }_{1/4}(x^2+3x+8)$

$-\frac{1}{2}{\log }_2(x+1)\ge +\frac{2}{4}{\log }_2\left(2\right)-\frac{1}{2}{\log }_2(x^2+3x+8)$

$-{\log }_2(x+1)\ge 1-{\log }_2(x^2+3x+8)$

${\log }_2(x^2+3x+8)-{\log }_2(x+1)\ge 1$

$\frac{(x^2+3x+8)}{(x+1)}\ge 2$

$(x+1)(x^2+3x+8)\ge 2{(x+1)}^2$

$(x+1)[x^2+3x+8-2(x+1\left)\right]\ge 0$

$(x+1)(x^2+x+6)\ge 0$

($x^2+x+6$ is also positive) $D<0$

Hence, $(x+1)\ge 0$

$xϵ[-1,\infty )$

Taking intersection we get $xϵ(-1,\infty )$