Question

Solution set of the inequality $(x-2{)}^{x^2-6x+8}>1$, where $x>2$ is

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

Answer 1

The correct option is A $(2,3)\cup (4,\infty )$

$(x-2{)}^{x^2-6x+8}>1$
For $x>2$
$\Rightarrow (x^2-6x+8){\log }_{10}(x-2)>0$
$\Rightarrow (x-2)(x-4){\log }_{10}(x-2)>0$

$\text{Case 1:}$
$(x-2)(x-4)>0\text{and}{\log }_{10}(x-2)>0$
$\Rightarrow x\in \left(\right(-\infty ,2)\cup (4,\infty \left)\right)\cap (3,\infty )$
$\Rightarrow x\in (4,\infty )\cdots \left(1\right)$

$\text{Case 2:}$
$\Rightarrow (x-2)(x-4)<0\text{and}{\log }_{10}(x-2)<0$
$\Rightarrow x\in (2,4)\cap (2,3)$
$\Rightarrow x\in (2,3)\cdots \left(2\right)$

Now, $\left(1\right)\cup \left(2\right)$
$\Rightarrow x\in (2,3)\cup (4,\infty )$