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

1. For $a\ne 0,1$ , if ${{a}^{x}}=b$ , then we say ${{\log }_{a}}b=x$ .
2. Observe the following facts:
   2.1) ${{\log }_{a}}b=1$ , if $a=b$.
   2.2) ${{\log }_{a}} b >1$ , if $a>1; a < b$ OR $a<1;a>b$.
   2.3) ${{\log }_{a}}b<1$ , if $a<1; a< b$ OR $a> 1; a>b$.
3. In other words, for $a\ne 0,1$ :
   3.1) ${{a}^{x}}={{a}^{y}}$ , if $x=y$.
   3.2) ${{a}^{x}}>{{a}^{y}}$ , if $a>1;x>y$ OR $a<1;x3.3)$ {{a}^{x}}<{{a}^{y}} $, if$ a<1 ; x>y $OR$ a > 1 ; x4) Convert the "log" into index (power) form and solve the inequation.
4. Multiplying / Dividing by a negative quantity reverses the sign of the inequality:
   If $a>b$ , then $a^n>b^n$ if $n>0$.

Complete step by step solution:
We know that ${{\log }_{a}}b\le 1$ , if $a<1;a\le b$ OR $a>1;a\ge b$ .
Let's say that $\dfrac{x+2}{x}=y;x\ne 0$ .
Since $0.2<1$ , we have the following case for ${{\log }_{0.2}}y\le 1$ :
$0.2\le y$
⇒ $0.2\le \dfrac{x+2}{x}$
Now, multiplying both sides by x, we get the following two cases:
CASE 1: If $x>0$ , then:
$0.2x\le x+2$
⇒ $-0.8x\le 2$
Dividing both sides by -0.8 (a negative number), we get:
⇒ $x\ge -\dfrac{2}{0.8}=-\dfrac{20}{8}=-\dfrac{5}{2}$
Since $x>0$ , the required values of x can be written as:
$x\in \left( 0,+\infty \right)$
CASE 2: If $x<0$ , then:
$0.2x\ge x+2$
⇒ $-0.8x\ge 2$
Dividing both sides by -0.8 (a negative number), we get:
⇒ $x\le -\dfrac{2}{0.8}=-\dfrac{20}{8}=-\dfrac{5}{2}$
Since $x<0$ , the required values of x can be written as:
$x\in \left( -\infty ,-\dfrac{5}{2} \right]$

∴ The correct answer is A. $\left( -\infty ,-\dfrac{5}{2} \right]\cup \left( 0,+\infty \right)$ .

Note:

1. Logarithm of 1 to any base (non-zero) is 0.
   ${{\log }_{a}}1=0$ , as ${{a}^{0}}=1;a\ne 0$
2. The following formulae are useful:
   $\log (mn)=\log m+\log n$
   $\log \left( \dfrac{m}{n} \right)=\log m-\log n$
   $\log {{a}^{x}}=x\log a$
3. Logarithms convert a complex mathematical calculation into simple additions and subtractions.