Question

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

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

Answer 1

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

${(x-2)}^{x^2-6x+8}>1$
For $x>2$
$(x^2-6x+8){\log }_{10}(x-2)>0$
$\Rightarrow (x-2)(x-4){\log }_{10}(x-2)>0$
$\Rightarrow (x-2)(x-4)>0,{\log }_{10}(x-2)>0$ or $(x-2)(x-4)<0,{\log }_{10}(x-2)<0$
$\Rightarrow x>4$
Therefore, $x\in (2,3)\cup (4,\infty )$
Ans: A