The solution set of log8⁡x2−4x+3≤1 is

log8⁡x2−4x+3≤1⇔x2−4x+3≤8 and x2−4x+3>0⇔x2−4x−5≤0 and x2−4x+3>0⇔(x−5)(x+1)≤0 and (x−1)(x−3)>0⇔−1≤x≤5 and x∉[1,3]⇔−1≤x<1 or 3<x≤5

The solution set of log8⁡x2−4x+3≤1 is

log8⁡x2−4x+3≤1⇔x2−4x+3≤8 and x2−4x+3>0⇔x2−4x−5≤0 and x2−4x+3>0⇔(x−5)(x+1)≤0 and (x−1)(x−3)>0⇔−1≤x≤5 and x∉[1,3]⇔−1≤x<1 or 3<x≤5

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The solution set of $\log_{8} (x^{2} - 4 x + 3) \leq 1$ is

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$(- 1, 1) \cup (3, 5)$

$[- 1, 1) \cup (3, 5]$

$[- 1, 1] \cup [3, 5]$

$(- 1, 1] \cup (3, 5)$

answer is B.

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Detailed Solution

$\begin{array}{l} \log_{8} (x^{2} - 4 x + 3) \leq 1 \\ \Leftrightarrow x^{2} - 4 x + 3 \leq 8 and x^{2} - 4 x + 3 > 0 \\ \Leftrightarrow x^{2} - 4 x - 5 \leq 0 and x^{2} - 4 x + 3 > 0 \\ \Leftrightarrow (x - 5) (x + 1) \leq 0 and (x - 1) (x - 3) > 0 \\ \Leftrightarrow - 1 \leq x \leq 5 and x \notin [1, 3] \\ \Leftrightarrow - 1 \leq x < 1 or 3 < x \leq 5 \end{array}$