Question

The set of real values of $x$ for which ${\log _{0.2}}\dfrac{{x + 2}}{x} \le 1$ is
A.$\left( { - \infty , - \dfrac{5}{2}} \right] \cup \left( {0, + \infty } \right)$
B.$\left[ {\dfrac{5}{2}, + \infty } \right)$
C.$\left( { - \infty ,0} \right) \cup \left( {\dfrac{8}{5}, + \infty } \right)$
D.None of these

Answer 1

Here in this question, we have to form the condition of the set of ranges for $x$. We will first use the basic criteria of the log function for always taking the function with the value greater than zero. Then we will solve the given equation to get a condition for the set of ranges of $x$. Then by combining all the conditions we will get the set of real values of $x$.

Complete step by step solution:
We know that the log function only takes the value of the function which is greater than zero. Therefore we get
$\dfrac{{x + 2}}{x} > 0$
From the above condition we can say that either $x > 0$ or $x + 2 > 0$. Therefore from this we get
$x > 0$……………………$\left( 1 \right)$
$x < \- 2$…………………$\left( 2 \right)$
Now simplifying the given equation, we get
${\log _{0.2}}\dfrac{{x + 2}}{x} \le 1$
We know this property of the log function that ${\log _a}b = \dfrac{{\log b}}{{\log a}}$. We can write this log function as
$\Rightarrow \dfrac{{\log \dfrac{{x + 2}}{x}}}{{\log 0.2}} \le 1$
Now we will simplify this equation to get the condition of the value of $x$. Therefore we get
$\Rightarrow \log \dfrac{{x + 2}}{x} \ge \log 0.2$
$\Rightarrow \log \dfrac{{x + 2}}{x} \ge \log \dfrac{1}{5}$
Now log is on the both side of the equation therefore it cancels out, we get
$\Rightarrow \dfrac{{x + 2}}{x} \ge \dfrac{1}{5}$
$\Rightarrow \left( {x + 2} \right)5 \ge x$
Now by solving this we will get
$\Rightarrow x \le \dfrac{{ - 5}}{2}$……………………$\left( 3 \right)$
Now by using the equations $\left( 1 \right)$, $\left( 2 \right)$ and $\left( 3 \right)$. We will get the set of ranges of $x$. Therefore we get
$x \in \left( { - \infty , - \dfrac{5}{2}} \right] \cup \left( {0, + \infty } \right)$
Hence, the set of real values of $x$ for which ${\log _{0.2}}\dfrac{{x + 2}}{x} \le 1$ is $\left( { - \infty , - \dfrac{5}{2}} \right] \cup \left( {0, + \infty } \right)$
So, option A is the correct option.

Note: We should take care of the equality and greater than sign while calculation. In this type of questions we have to make the conditions by the given equations only and then we will combine those equations. We should know that we cannot find the log function of 0 or any negative number, the number should always be greater than zero.
Some of the basic properties of the log functions are:
1.$\log a + \log b = \log ab$
2.$\log a - \log b = \log \dfrac{a}{b}$
3.${\log _a}b = \dfrac{{\log b}}{{\log a}}$