Question

The solution set of the inequality ${\log }_3\left(\right(x+2\left)\right(x+4\left)\right)+{\log }_{1/3}(x+2)<\frac{1}{2}{\log }_{\sqrt{3}}7$ is

• A. $(-2,-1)$
• B. $(-2,3)$
• C. $(-1,3)$
• D. $(3,\infty )$

Answer 1

The correct option is B $(-2,3)$

${\log }_3\left[\right(x+2\left)\right(x+4\left)\right]+{\log }_{1/3}(x+2)<\frac{1}{2}{\log }_{\sqrt{3}}7$
$\Rightarrow {\log }_3(x+2)(x+4)-{\log }_3(x+2)<{\log }_{3}7$
$\Rightarrow {\log }_3(x+4)<{\log }_{3}7$
$\Rightarrow x+4<7$
$\Rightarrow x<3\cdots \left(1\right)$

For domain $\to (x+2)(x+4)>0$
$\Rightarrow x\in (-\infty ,-4)\cup (-2,\infty )\cdots \left(2\right)$
and $x+2>0\cdots \left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right),$
$x\in (-2,3)$