Question

The solution set of the inequality ${\log }_{1/3}(x^2-6x+18)-2{\log }_{1/3}(x-4)<0$ is

• A. $(3,4)$
• B. $(4,\infty )$
• C. $(2,4)$
• D. $(7,\infty )$

Answer 1

Answer: (B)

$(4,\infty )$

${\log }_{1/3}(x^2-6x+18)-2{\log }_{1/3}(x-4)<0$
Above equation is valid when $x^2-6x+18>0\Rightarrow x\in R$
and $x-4>0\Rightarrow x>4$
$\Rightarrow x>4$ ....(1)
Now
${\log }_{1/3}(x^2-6x+18)-2{\log }_{1/3}(x-4)<0$
$\Rightarrow {\log }_{1/3}\frac{(x^2-6x+18)}{{(x-4)}^2}<0[\because m\log x=\log x^m\text{\&}\log a-\log b=\log \frac{a}{b}]$
$\Rightarrow \frac{(x^2-6x+18)}{{(x-4)}^2}>1[\because$ base of log is less than 1]
$\Rightarrow \frac{2x+2}{{(x-4)}^2}>0$
$\Rightarrow 2x+2>0$
$\Rightarrow x>-1$ ...(2)

From (1) & (2), we get $x>4$
Hence, option B.