Question

Find the largest integral value of $x$ satisfying the inequality ${\log }_{0.09}\left(x^2+2x\right)\ge {\log }_{0.3}\sqrt{x+2}$.

Answer 1

Answer: 1

${\log }_{0.09}\left(x^2+2x\right)$ is defined, when $x^2+2x>0\Rightarrow x(x+2)>0$
$\Rightarrow x\in (-\infty ,-2)Y(0,\infty )$ .....(1)
And ${\log }_{0.3}\sqrt{x+2}$ is defined, when $x+2>0\Rightarrow x\in (-2,\infty )$ ....(2)

Also ${\log }_{0.09}\left(x^2+2x\right)\ge {\log }_{0.3}\sqrt{x+2}$
$\Rightarrow \frac{1}{2}{\log }_{0.3}\left(x^2+2x\right)\ge {\log }_{0.3}\sqrt{x+2}$
$\Rightarrow {\log }_{0.3}\left(x^2+2x\right)\ge {\log }_{0.3}(x+2)$
$\Rightarrow x^2+2x\le x+2$
$\Rightarrow x^2+x-2\le 0\Rightarrow (x+2)(x-1)\le 0$
$\Rightarrow x\in [-2,1]$
$\therefore$ From (1), (2) and (3), we get $x\in (0,1]$.
Hence largest integral value of $x=1$. Ans.]