Question

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

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

Answer 1

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

${\log }_3(x+2)(x+4)+{\log }_{\frac{1}{3}}(x+2)<\frac{1}{2}{\log }_{\sqrt{3}}7.......\left(1\right)$
above equation (1) is valid when,
$(x+2)(x+4)>0,x+2>0$
$\Rightarrow x>-2$ .......$\left(2\right)$
Now (1) can be written as
${\log }_3(x+2)(x+4)-{\log }_3(x+2)<\frac{\left(\log 7\right)}{\left(\log 3\right)}[\because {\log }_{1/a}b=-{\log }_{a}b,{\log }_{a}b=\frac{\log b}{\log a},{\log }_{a^b}c=\frac{1}{b}{\log }_{a}c]$
$\Rightarrow {\log }_3(x+2)+{\log }_3(x+4)-{\log }_3(x+2)<{\log }_{3}7[\because \log ab=\log a+\log b,\log a-\log b=\log \frac{a}{b}]$

$\Rightarrow {\log }_3(x+4)<{\log }_{3}7$

$\Rightarrow x+4<7orx<3$ ....(3)
From (2) & (3) , we get
$x\in (-2,3)$

Ans: B