Question

Let $a>0,a\ne 1$ and $x>0$
(i) ${\log }_{a}x>0,a>1\iff x>0,a>1$
(ii)${\log }_{a}x>0,0<a<1\iff 0<x<1,0<a<1$
(iii) ${\log }_{a}x>0,a>1\iff 0<x<1,a>1$
(iv) ${\log }_{a}x<0,0<a<1\iff x>1,0<a<1$

What is the solution set of ${\log }_3\frac{5-2x}{12x-8}+{\log }_{1/3}x\le 0$?

• A. $(\frac{5}{12},\frac{2}{3})$
• B. $(\frac{5}{12},\frac{1}{2})$
• C. $(\frac{5}{12},\frac{1}{3})$
• D. $(\frac{5}{12},\frac{-1}{2})$

Answer 1

Answer: (B)

$(\frac{5}{12},\frac{1}{2})$

${\log }_3\left(\frac{5-2x}{12x-8}\right)+{\log }_{1/3}x\le 0$

$\Rightarrow {\log }_3\left(\frac{5-2x}{12x-8}\right)-1{\log }_{3}x\le 0$

$\Rightarrow {\log }_3\left(\frac{5-2x}{12x^2-8x}\right)\le 0$

$\Rightarrow 0<\frac{5-2x}{12x^2-8x}<1$

Solving the inequality we get:

$x\in (\frac{5}{12},\frac{1}{2})$