Question

Solve the following inequality:
${\left(\frac{2}{5}\right)}^{{\log }_{0.25}(x^2+5x+8)}⩽2.5$

Answer 1

${\left(\frac{2}{5}\right)}^{\frac{\log (x^2+5x+8)}{-\log 4}}\le 2.5$ and $x^2+5x+8>0x\in R$

${\left(\frac{5}{2}\right)}^{{\log }_4(x^2+5x+8)}\le 2.5{\left(\frac{5}{2}\right)}^{{\log }_4(x^2+5x+8)}\le {\left(2.5\right)}^1{\log }_4(x^2+5x+8)\le 1\log (x^2+5x+8)\le \log 4(x^2+5x+8)\le 4x^2+5x+4\le 0(x+1)(x+4)\le 0x\in [-4,-1]$

Taking intersection, we get $x\in [-4,-1]$