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Question Number 100666 by bobhans last updated on 28/Jun/20

$find solution set of inequality$ ${(\log_{2} x - 2)}^{3 x - 1} < {(\log_{2} x - 2)}^{3 - x}$
Commented by Rasheed.Sindhi last updated on 28/Jun/20

${(\log_{2} x - 2)}^{3 x - 1} < {(\log_{2} x - 2)}^{3 - x}$ $\log_{2} x - 2 \neq 0, 1 \Rightarrow 3 x - 1 < 3 - x$ $\Rightarrow 4 x < 4 \Rightarrow x < 1$
Commented by bramlex last updated on 28/Jun/20

$i t h i n k n o t c o r r e c t s i r$
Commented by Rasheed.Sindhi last updated on 28/Jun/20

${Y o u}^{'} r e r i g h t s i r!$
	Answered by bramlex last updated on 28/Jun/20

	$\Leftrightarrow (\log_{2} x - 2 - 1) (3 x - 1 - (3 - x)) < 0$ $(\log_{2} x - 3) (4 x - 4) < 0$ $4 (\log_{2} x - 3) (x - 1) < 0$ $c a s e 1 \Rightarrow \log_{2} x - 3 < 0 \land x - 1 > 0$ $x < 8 \land x > 1 \Rightarrow 1 < x < 8$ $c a s e 2 \Rightarrow \log_{2} x - 3 > 0 \land x - 1 < 0$ $x > 8 \land x < 1 \Rightarrow x = \emptyset$ $s o l u t i o n (1) \cup (2) \Rightarrow 1 < x < 8$
	Commented by bemath last updated on 28/Jun/20

	$sir bramlex it should be (\log_{2} x - 2 - 1)$ $n o t (\log_{2} x - 2 + 1)$
	Commented by bramlex last updated on 28/Jun/20

	$o o y e s . . y o u r a r e r i g h t$
	Answered by 1549442205 last updated on 28/Jun/20
	$the condition for the given inequality is defined as { ((x>0)),((log_2 x−2>0)) :} ⇔x>4.Then The given inequality is equivalent to [log_2 x−2−1)[(3x−1)−(3−x)]<0 ⇔(log_2 x−3)(4x−4)<0⇔(log_2 x−3)(x−1)<0 ⇔[_( { ((log_2 x−3<0)),((x−1>0)) :} ⇔ { ((4<x<8)),((x>1)) :} ⇔4<x<8) ^( { ((log_2 x−3>0)),((x−1<0)) :} ⇔ { ((x>8)),((x<1)) :} ⇒has no solutions) Thus,solution set of the inequality is the interval (4,8)$
	$the condition for the given inequality$ $is defined as {\begin{cases} x > 0 \\ \log_{2} x - 2 > 0 \end{cases} \Leftrightarrow x > 4 . Then$ $The given inequality is equivalent to$ $[\log_{2} x - 2 - 1) [(3 x - 1) - (3 - x)] < 0$ $\Leftrightarrow (\log_{2} x - 3) (4 x - 4) < 0 \Leftrightarrow (\log_{2} x - 3) (x - 1) < 0$ $\Leftrightarrow [_{{\begin{cases} \log_{2} x - 3 < 0 \\ x - 1 > 0 \end{cases} \Leftrightarrow {\begin{cases} 4 < x < 8 \\ x > 1 \end{cases} \Leftrightarrow 4 < x < 8}^{{\begin{cases} \log_{2} x - 3 > 0 \\ x - 1 < 0 \end{cases} \Leftrightarrow {\begin{cases} x > 8 \\ x < 1 \end{cases} \Rightarrow has no solutions}$ $Thus, solution set of the inequality is$ $the interval (4, 8)$
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