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Q.
Find the largest integral value of $x$ satisfying the inequality $\log _{0.09}\left(x^2+2 x\right) \geq \log _{0.3} \sqrt{x+2}$.
Complex Numbers and Quadratic Equations
Solution:
$\log _{0.09}\left(x^2+2 x\right)$ is defined, when $x^2+2 x>0 \Rightarrow x(x+2)>0$
$\Rightarrow x \in(-\infty,-2) Y(0, \infty)$ .....(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+2 x\right) \geq \log _{0.3} \sqrt{x+2}$
$\Rightarrow \frac{1}{2} \log _{0.3}\left(x^2+2 x\right) \geq \log _{0.3} \sqrt{x+2} $
$\Rightarrow \log _{0.3}\left(x^2+2 x\right) \geq \log _{0.3}(x+2) $
$\Rightarrow x^2+2 x \leq x+2$
$\Rightarrow x^2+x-2 \leq 0 \Rightarrow(x+2)(x-1) \leq 0$
$\Rightarrow x \in[-2,1]$
$\therefore $ From (1), (2) and (3), we get $x \in(0,1]$.
Hence largest integral value of $x=1$. Ans.]
