Question

The values of $x$ for which ${\log }_3(x+2)(x+4)+{\log }_{1/3}(x+2)<\frac{1}{2}{\log }_{\sqrt{3}}7$ holds, is

• A. $-4<x<3$
• B. $x<3$
• C. $x>3$
• D. $-2<x<3$

Answer 1

The correct option is D $-2<x<3$

Clearly, $(x+2)(x+4)>0$ and $x+2>0$
$\Rightarrow x>-2\cdots \left(1\right)$

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

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