The set of real values of x for which log_{0. | vedclass

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(A) Given the inequality: $\log_{0.2} \left( \frac{x + 2}{x} \right) \le 1$ ... $(i)$For the logarithm to be defined,the argument must be positive:$\f

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

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(A) Given the inequality: $\log_{0.2} \left( \frac{x + 2}{x} \right) \le 1$ ... $(i)$For the logarithm to be defined,the argument must be positive:$\frac{x + 2}{x} > 0 \Rightarrow x \in (-\infty, -2) \cup (0, \infty)$Since the base $0.2 $\frac{x + 2}{x} \ge (0.2)^1$$\frac{x + 2}{x} \ge 0.2$$\frac{x + 2}{x} - 0.2 \ge 0$$\frac{x + 2 - 0.2x}{x} \ge 0$$\frac{0.8x + 2}{x} \ge 0$$\frac{4x + 10}{5x} \ge 0$$\frac{2(2x + 5)}{5x} \ge 0$Using the wavy curve method,the solution to $\frac{2x + 5}{x} \ge 0$ is $x \in (-\infty, -2.5] \cup (0, \infty)$.Combining this with the domain condition $x \in (-\infty, -2) \cup (0, \infty)$,the intersection is $x \in (-\infty, -2.5] \cup (0, \infty)$.

The set of real values of $x$ for which $\log_{0.2} \left( \frac{x + 2}{x} \right) \le 1$ is

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