Question

The set of values of $x$ satisfying the inequality $(\frac{7}{9^x-2}\ge \frac{2}{3^x-1})$ is

• A. $x\in (-{\log }_{3}2,0)\cup ({\log }_3\sqrt{2},1]$
• B. $x\in [-{\log }_{3}2,0)\cup ({\log }_3\sqrt{2},1)$
• C. $x\in [-{\log }_{3}2,0)\cup [{\log }_3\sqrt{2},1]$
• D. None of these

Answer 1

Answer: (B)

$x\in [-{\log }_{3}2,0)\cup ({\log }_3\sqrt{2},1)$

Given,

$\frac{7}{9^x-2}\ge \frac{2}{3^x-1}$

Above inequality is not defined when $3^x=1\Rightarrow x=0$ .
Also, it is not defined when $9^x=2$
$\Rightarrow 3^{2x}=2$
$\Rightarrow 2x={\log }_{3}2$
$\Rightarrow x={\log }_3\sqrt{2}$
So, the above inequality is not defined at $x=0$ and $x={\log }_3\sqrt{2}$
Now, for solving the inequality, put $3^x=t$
$\frac{7}{t^2-2}\ge \frac{2}{t-1}$
$\Rightarrow 2t^2-7t+3\le 0$
$\Rightarrow (t-3)(2t-1)\le 0$
$\Rightarrow \frac{1}{2}\le t\le 3$
$\Rightarrow \frac{1}{2}\le 3^x\le 3$
$\Rightarrow -{\log }_{3}2\le x\le 1$

So, $x\in [-{\log }_{3}2,1]-\{0,{\log }_3\sqrt{2}\}$