Question

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

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

Answer 1

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

For the inequality to be defined,
$(x+2)(x+4)>0\text{and}x+2>0$
$\Rightarrow x\in (-\infty ,-4)\cup (-2,\infty )\text{and}x\in (-2,\infty )$
So, $x\in (-2,\infty )$

Now,
${\log }_3(x+2)(x+4)+{\log }_{1/3}(x+2)<\frac{1}{2}{\log }_{\sqrt{3}}7$
$\Rightarrow {\log }_3(x+2)(x+4)-{\log }_3(x+2)<\frac{2}{2}{\log }_{3}7\Rightarrow {\log }_3(x+2)(x+4)-{\log }_3(x+2)-{\log }_{3}7<0$
$\Rightarrow {\log }_3\frac{(x+2)(x+4)}{7(x+2)}<0$
As the base of log function is greater than $1$, the inequality sign will remain same.
$\Rightarrow \frac{(x+2)(x+4)}{7(x+2)}<1\Rightarrow \frac{x+4}{7}<1[\because x>-2\Rightarrow x+2\ne 0]\Rightarrow x<3$
Hence, $x\in (-2,3)$