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 $(-2,3)$

${\log }_3(x+2)(x+4)-{\log }_3(x+2)<{\log }_{3}7$
${\log }_3(x+4)<{\log }_{3}7$
$x+4<7$
$x<3...\left(1\right)$
For domain $\to (x+2)(x+4)>0$ ,
$x>-2orx<-4...\left(2\right)$

$x+2>0...\left(3\right)$

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