Question

The complete solution set of the inequality
${\log }_{1/3}(x+1)>{\log }_3(3-x)$ is

• A. $x\in (1-\sqrt{3},3)$
• B. $x\in (-1,1-\sqrt{3})\cup (1+\sqrt{3},3)$
• C. $x\in (-1,1+\sqrt{3})$
• D. $x\in (-1,1)\cup (1+\sqrt{3},3)$

Answer 1

The correct option is B $x\in (-1,1-\sqrt{3})\cup (1+\sqrt{3},3)$

${\log }_{1/3}(x+1)>{\log }_3(3-x)$
For $\log$ to be defined
$(x+1)>0$ and $(3-x)>0$
$\Rightarrow x>-1$ and $x<3$
$\Rightarrow x\in (-1,3)\cdots \left(1\right)$

Now, $-{\log }_3(x+1)>{\log }_3(3-x)$
$\Rightarrow {\log }_3(x+1{)}^{-1}>{\log }_3(3-x)$
$\Rightarrow \frac{1}{x+1}>(3-x)$
$\Rightarrow \frac{x^2-2x-2}{x+1}>0$
$\Rightarrow x\in (-1,1-\sqrt{3})\cup (1+\sqrt{3},\infty )\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$x\in (-1,1-\sqrt{3})\cup (1+\sqrt{3},3)$