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

(2, 3) \cup (4, \infty)

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(2, \infty)

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(- 3, \infty)

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(1, 2)

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Solution

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

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