The set of real values of x for which {log _{ | vedclass

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Question

(a) ${\log _{0.2}}{{x + 2} \over x} \le 1$&hellip;..$(i)$

For log to be defined, ${{x + 2} \over x} > 0$ $ \Rightarrow $$x > 0$ or $x < -

Vedclass · Accepted Answer

$\left( { - \infty ,\,\, - {5 \over 2}} \right] \cup (0, + \infty )$

Vedclass · Answer

(a) ${\log _{0.2}}{{x + 2} \over x} \le 1$…..$(i)$ For log to be defined, ${{x + 2} \over x} > 0$ $ \Rightarrow $$x > 0$ or $x < - 2$ Now from $(i),$ ${\log _{0.2}}{{x + 2} \over x} \le {\log _{0.2}}0.2$ $ \Rightarrow $${{x + 2} \over x} \ge 0.2$ …..$(ii)$ Case $(i)$ $x > 0$ From $(ii),$ $x + 2 \ge 0.2x$ $ \Rightarrow $ $0.8x \ge - 2$ $ \Rightarrow $$x \ge - {5 \over 2}$. Case $(ii)$ $x < - 2$ From $(ii),$ $x + 2 \le 0.2x$$ \Rightarrow $$0.8x \le - 2$$ \Rightarrow $$x \le - {5 \over 2}$ $ \Rightarrow $$x \in (0,\,\infty )\, \cup \,\left( { - \infty ,\, - {5 \over 2}} ight]$; $ herefore x \in \left( { - \infty ,\, - {5 \over 2}} ight] \cup (0,\,\infty )$.

The set of real values of $x$ for which ${\log _{0.2}}{{x + 2} \over x} \le 1$ is

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