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If $\log_{3} (x^{2} - 5 x + 7) < 0$ , $x$ then lies in the interval

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$(- \infty, 2) \cup (3, \infty)$

$(2, 3)$

$(- \infty, 2) \cup (2, \infty)$

$[2, 3]$

answer is B.

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Detailed Solution

$\log_{3} (x^{2} - 5 x + 7) < 0$

$\log_{3} (x^{2} - 5 x + 7) < \log_{3}^{1}$

$\Rightarrow x^{2} - 5 x + 7 < 1$

$x^{2} - 5 x + 6 < 0$

$x^{2} - 3 x - 2 x + 6 < 0$

$x (x - 3) - 2 (x - 3) < 0$

$(x - 3) (x - 2) < 0$

Hence $x \in (2, 3)$

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