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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If $x = \log_a (bc)$, $y = \log_b (ca)$, and $z = \log_c (ab)$, then the value of $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z}$ is

The solution to the equation ${\log _7}{\log _5}(\sqrt {{x^2} + 5 + x} ) = 0$ is:

The number of solutions of the equation $\log _{2}\left(x^{2}+2 x-1\right)=1$ is

The value of $\log _4 18$ is

If $3^x = 4^{x-1}$, then find the value of $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module