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Question Number 100666 by bobhans last updated on 28/Jun/20
find solution set of inequality (log _2 x −2)^(3x−1) < (log _2 x−2)^(3−x)
$$\mathrm{find}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{inequality} \\ $$$$\left(\mathrm{log}\:_{\mathrm{2}} {x}\:−\mathrm{2}\right)^{\mathrm{3}{x}−\mathrm{1}} \:<\:\left(\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{2}\right)^{\mathrm{3}−{x}} \\ $$
Commented by Rasheed.Sindhi last updated on 28/Jun/20
(log _2 x −2)^(3x−1) < (log _2 x−2)^(3−x) log _2 x −2≠0,1⇒3x−1<3−x ⇒4x<4⇒x<1
$$\left(\mathrm{log}\:_{\mathrm{2}} {x}\:−\mathrm{2}\right)^{\mathrm{3}{x}−\mathrm{1}} \:<\:\left(\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{2}\right)^{\mathrm{3}−{x}} \\ $$$$\mathrm{log}\:_{\mathrm{2}} {x}\:−\mathrm{2}\neq\mathrm{0},\mathrm{1}\Rightarrow\mathrm{3}{x}−\mathrm{1}<\mathrm{3}−{x} \\ $$$$\Rightarrow\mathrm{4}{x}<\mathrm{4}\Rightarrow{x}<\mathrm{1} \\ $$
Commented by bramlex last updated on 28/Jun/20
i think not correct sir
$${i}\:{think}\:{not}\:{correct}\:{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 28/Jun/20
You′re right sir!
$${You}'{re}\:{right}\:{sir}! \\ $$
Answered by bramlex last updated on 28/Jun/20
⇔(log _2 x−2−1)(3x−1−(3−x))<0 (log _2 x−3)(4x−4)<0 4(log _2 x−3)(x−1) <0 case 1 ⇒ log _2 x−3<0 ∧ x−1>0 x<8 ∧x>1 ⇒1<x<8 case 2 ⇒ log _2 x−3>0 ∧ x−1<0 x >8 ∧ x<1 ⇒x = ∅ solution (1)∪(2) ⇒ 1 < x < 8
$$\Leftrightarrow\left(\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{2}−\mathrm{1}\right)\left(\mathrm{3}{x}−\mathrm{1}−\left(\mathrm{3}−{x}\right)\right)<\mathrm{0} \\ $$$$\left(\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{3}\right)\left(\mathrm{4}{x}−\mathrm{4}\right)<\mathrm{0}\: \\ $$$$\mathrm{4}\left(\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{3}\right)\left({x}−\mathrm{1}\right)\:<\mathrm{0} \\ $$$${case}\:\mathrm{1}\:\Rightarrow\:\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{3}<\mathrm{0}\:\wedge\:{x}−\mathrm{1}>\mathrm{0} \\ $$$${x}<\mathrm{8}\:\wedge{x}>\mathrm{1}\:\Rightarrow\mathrm{1}<{x}<\mathrm{8} \\ $$$${case}\:\mathrm{2}\:\Rightarrow\:\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{3}>\mathrm{0}\:\wedge\:{x}−\mathrm{1}<\mathrm{0} \\ $$$${x}\:>\mathrm{8}\:\wedge\:{x}<\mathrm{1}\:\Rightarrow{x}\:=\:\varnothing \\ $$$${solution}\:\left(\mathrm{1}\right)\cup\left(\mathrm{2}\right)\:\Rightarrow\:\mathrm{1}\:<\:{x}\:<\:\mathrm{8}\: \\ $$
Commented by bemath last updated on 28/Jun/20
sir bramlex it should be (log _2 x−2−1) not (log _2 x−2+1)
$$\mathrm{sir}\:\mathrm{bramlex}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\left(\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{2}−\mathrm{1}\right) \\ $$$${not}\:\left(\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{2}+\mathrm{1}\right) \\ $$
Commented by bramlex last updated on 28/Jun/20
oo yes..your are right
$${oo}\:{yes}..{your}\:{are}\:{right} \\ $$
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)
$$ \\ $$$$\mathrm{the}\:\mathrm{condition}\:\mathrm{for}\:\mathrm{the}\:\mathrm{given}\:\mathrm{inequality}\: \\ $$$$\mathrm{is}\:\mathrm{defined}\:\mathrm{as\begin{cases}{\mathrm{x}>\mathrm{0}}\\{\mathrm{log}_{\mathrm{2}} \mathrm{x}−\mathrm{2}>\mathrm{0}}\end{cases}}\:\:\Leftrightarrow\mathrm{x}>\mathrm{4}.\mathrm{Then} \\ $$$$\mathrm{The}\:\mathrm{given}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to} \\ $$$$\left[\mathrm{log}_{\mathrm{2}} \mathrm{x}−\mathrm{2}−\mathrm{1}\right)\left[\left(\mathrm{3x}−\mathrm{1}\right)−\left(\mathrm{3}−\mathrm{x}\right)\right]<\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{log}_{\mathrm{2}} \mathrm{x}−\mathrm{3}\right)\left(\mathrm{4x}−\mathrm{4}\right)<\mathrm{0}\Leftrightarrow\left(\mathrm{log}_{\mathrm{2}} \mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}−\mathrm{1}\right)<\mathrm{0} \\ $$$$\Leftrightarrow\left[_{\begin{cases}{\mathrm{log}_{\mathrm{2}} \mathrm{x}−\mathrm{3}<\mathrm{0}}\\{\mathrm{x}−\mathrm{1}>\mathrm{0}}\end{cases}\:\:\Leftrightarrow\begin{cases}{\mathrm{4}<\mathrm{x}<\mathrm{8}}\\{\mathrm{x}>\mathrm{1}}\end{cases}\:\:\:\Leftrightarrow\mathrm{4}<\mathrm{x}<\mathrm{8}} ^{\begin{cases}{\mathrm{log}_{\mathrm{2}} \mathrm{x}−\mathrm{3}>\mathrm{0}}\\{\mathrm{x}−\mathrm{1}<\mathrm{0}}\end{cases}\:\:\Leftrightarrow\begin{cases}{\mathrm{x}>\mathrm{8}}\\{\mathrm{x}<\mathrm{1}}\end{cases}\:\:\:\Rightarrow\mathrm{has}\:\mathrm{no}\:\mathrm{solutions}} \right. \\ $$$$\mathrm{Thus},\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{interval}\:\left(\mathrm{4},\mathrm{8}\right) \\ $$$$ \\ $$

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