Question

The solution set of the inequality $3^{{\log }_3\sqrt{x-1}}<3^{{\log }_3(x-6)}+3$ is

• A. $(5,\infty )$
• B. $(6,\infty )$
• C. $(1,2)\cup (3,5)$
• D. $(-\infty ,2)\cup (5,\infty )$

Answer 1

The correct option is B $(6,\infty )$

$3^{{\log }_3\sqrt{x-1}}<3^{{\log }_3(x-6)}+3$
Clearly, $x>6$
$\Rightarrow \sqrt{x-1}<(x-6)+3$
$\Rightarrow \sqrt{x-1}<x-3$
Squaring both sides, we get
$(x-1)<(x-3{)}^2$
$\Rightarrow (x-1)<x^2-6x+9$
$\Rightarrow x^2-7x+10>0$
$\Rightarrow (x-5)(x-2)>0$
$\Rightarrow x\in (-\infty ,2)\cup (5,\infty )$
But $x>6$
$\therefore x\in (6,\infty )$