Question

If $x$ satisfies the equation ${\log }_{1/3}(x-1)+{\log }_{1/3}(x+1)+{\log }_{\sqrt{3}}(5-x)<1$, then $x$ lies in the interval

• A. $(1,3)$
• B. $(2,5)$
• C. $(1,2)$
• D. $(4,\infty )$

Answer 1

The correct option is B $(2,5)$

${\log }_{1/3}(x-1)+{\log }_{1/3}(x+1)+{\log }_{\sqrt{3}}(5-x)$ is valid when $x-1>0,x+1>0,5-x>0$
$\Rightarrow 1<x<5$ ...(1)
${\log }_{1/3}(x-1)+{\log }_{1/3}(x+1)+{\log }_{\sqrt{3}}(5-x)<1$
$\Rightarrow -{\log }_3(x-1)-{\log }_3(x+1)+2{\log }_3(5-x)<1[\because {\log }_{1/a}b=-{\log }_{a}b]$
$\Rightarrow {\log }_3\frac{{(5-x)}^2}{(x-1)(x+1)}<1$
$\Rightarrow \frac{{(5-x)}^2}{(x-1)(x+1)}<3$
$\Rightarrow x^2-10x+25<3x^2-3$
$\Rightarrow x^2+5x-14>0$
$\Rightarrow x<-7$ and $x>2$ ...(2)
From (1) & (2), we get $x\in (2,5)$