The set of real values of x satisfying the in | vedclass

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Question

(A) Given the inequality $\log_{0.2} \frac{x + 2}{x} \le 1$. Since the base $0.2 < 1$,the inequality sign reverses when removing the logarithm: $\frac

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the inequality $\log_{0.2} \frac{x + 2}{x} \le 1$. Since the base $0.2 $\frac{x + 2}{x} \ge (0.2)^1$ $\frac{x + 2}{x} \ge \frac{1}{5}$ $\frac{x + 2}{x} - \frac{1}{5} \ge 0$ $\frac{5(x + 2) - x}{5x} \ge 0$ $\frac{4x + 10}{5x} \ge 0$ $\frac{2x + 5}{x} \ge 0$ Also,the domain of the logarithm requires $\frac{x + 2}{x} > 0$,which implies $x \in ( - \infty, - 2 ) \cup (0, + \infty )$. Solving $\frac{2x + 5}{x} \ge 0$ gives $x \in ( - \infty, - \frac{5}{2} ] \cup (0, + \infty )$. Taking the intersection with the domain,we get $x \in ( - \infty, - \frac{5}{2} ] \cup (0, + \infty )$.

The set of real values of $x$ satisfying the inequality $\log_{0.2} \frac{x + 2}{x} \le 1$ is:

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