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

Hint:
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.
5) 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.