Question

Solve the inequality: ${\log }_3(x+2)(x+4)+{\log }_{1/3}(x+2)<\frac{1}{2}{\log }_{\sqrt{3}}7$

Answer 1

${\log }_3(x+2)(x+4)+{\log }_{\frac{1}{3}}(x+2)<\frac{1}{2}{\log }_{\sqrt{3}}7$

$\Rightarrow {\log }_3(x+2)(x+4)-{\log }_3(x+2)<\frac{2}{2}{\log }_{3}7$ ($\because {\log }_{a^b}x=\frac{1}{b}{\log }_{a}x$ )

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

$\Rightarrow x+4<7$ ($\because {\log }_{a}x<{\log }_{a}y;a>1\Rightarrow x<y$ )

$\Rightarrow x<3$ ...(1)

But log is defined when
$x+2>0$

$\Rightarrow x>-2$ ...(2)

$\therefore (x+2)(x+4)>0$

$\Rightarrow (-\infty ,-4)\cup (-2,\infty )$ ...(3)

From (1), (2) and (3)

$x\in (-2,3)$