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Question Number 87194 by john santu last updated on 03/Apr/20
find the solution of ((∣ log_2 (x)+2∣)/(x−3)) < 2
$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\: \\ $$$$\frac{\mid\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2}\mid}{\mathrm{x}−\mathrm{3}}\:<\:\mathrm{2}\: \\ $$
Commented by TANMAY PANACEA. last updated on 03/Apr/20
is it (2+log_2 x) or log_2 (x+2)
$${is}\:{it}\:\left(\mathrm{2}+{log}_{\mathrm{2}} {x}\right)\:\:{or}\:{log}_{\mathrm{2}} \left({x}+\mathrm{2}\right) \\ $$
Commented by john santu last updated on 03/Apr/20
2+ log_2 (x) sir
$$\mathrm{2}+\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)\:\mathrm{sir} \\ $$
Commented by john santu last updated on 03/Apr/20
i can^′ t answer that
$$\mathrm{i}\:\mathrm{can}^{'} \mathrm{t}\:\mathrm{answer}\:\mathrm{that} \\ $$
Commented by john santu last updated on 03/Apr/20
the choice answer is a) x < 2^3 or x > 2^8 b) 3 < x < 8 c) x < 3 or x > 8 d) 2^3 < x < 2^8 e) 0 < x < 2^3 or x > 2^8
$$\mathrm{the}\:\mathrm{choice}\:\mathrm{answer}\:\mathrm{is}\: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{x}\:<\:\mathrm{2}^{\mathrm{3}} \:\mathrm{or}\:\mathrm{x}\:>\:\mathrm{2}^{\mathrm{8}} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{3}\:<\:\mathrm{x}\:<\:\mathrm{8} \\ $$$$\left.\mathrm{c}\right)\:\mathrm{x}\:<\:\mathrm{3}\:\mathrm{or}\:\mathrm{x}\:>\:\mathrm{8} \\ $$$$\left.\mathrm{d}\right)\:\mathrm{2}^{\mathrm{3}} \:<\:\mathrm{x}\:<\:\mathrm{2}^{\mathrm{8}} \\ $$$$\left.\mathrm{e}\right)\:\mathrm{0}\:<\:\mathrm{x}\:<\:\mathrm{2}^{\mathrm{3}} \:\mathrm{or}\:\mathrm{x}\:>\:\mathrm{2}^{\mathrm{8}} \\ $$
Commented by jagoll last updated on 03/Apr/20
may be the question is ∣log_2 (((x+2)/(x−3)))∣ < 2
$$\mathrm{may}\:\mathrm{be}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\: \\ $$$$\mid\mathrm{log}_{\mathrm{2}} \:\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{3}}\right)\mid\:<\:\mathrm{2} \\ $$
Commented by jagoll last updated on 03/Apr/20
−2 < log_2 (((x+2)/(x−3))) < 2 2^(−2) < ((x+2)/(x−3)) < 2^2 ⇒ { ((((x+2)/(x−3)) > (1/4))),((((x+2)/(x−3)) < 4)) :}
$$−\mathrm{2}\:<\:\mathrm{log}_{\mathrm{2}} \:\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{3}}\right)\:<\:\mathrm{2} \\ $$$$\mathrm{2}^{−\mathrm{2}} \:<\:\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{3}}\:<\:\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow\:\begin{cases}{\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{3}}\:>\:\frac{\mathrm{1}}{\mathrm{4}}}\\{\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{3}}\:<\:\mathrm{4}}\end{cases}\: \\ $$
Answered by TANMAY PANACEA. last updated on 03/Apr/20
((∣2+log_2 x∣)/(x−3))−2<0 1)x≠3 and x≠0 assume x=2^k k>0 ((∣2+k∣)/(2^k −3))−2 =((2+k)/(2^k −3))−2 when k=1 ((2+1)/(2−3))−2<0 when k=2 ((2+2)/(4−3))−2>0 (note here ) when k=3 ((2+3)/(8−3))−2<0 k=4 ((2+4)/(16−3))−2<0
$$\frac{\mid\mathrm{2}+{log}_{\mathrm{2}} {x}\mid}{{x}−\mathrm{3}}−\mathrm{2}<\mathrm{0} \\ $$$$\left.\mathrm{1}\right){x}\neq\mathrm{3}\:\:\:{and}\:{x}\neq\mathrm{0} \\ $$$${assume}\:{x}=\mathrm{2}^{{k}} \:\:\:{k}>\mathrm{0} \\ $$$$\frac{\mid\mathrm{2}+{k}\mid}{\mathrm{2}^{{k}} −\mathrm{3}}−\mathrm{2} \\ $$$$=\frac{\mathrm{2}+{k}}{\mathrm{2}^{{k}} −\mathrm{3}}−\mathrm{2} \\ $$$${when}\:{k}=\mathrm{1} \\ $$$$\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}−\mathrm{3}}−\mathrm{2}<\mathrm{0} \\ $$$${when}\:{k}=\mathrm{2} \\ $$$$\frac{\mathrm{2}+\mathrm{2}}{\mathrm{4}−\mathrm{3}}−\mathrm{2}>\mathrm{0}\:\:\left({note}\:{here}\:\right) \\ $$$${when}\:{k}=\mathrm{3} \\ $$$$\frac{\mathrm{2}+\mathrm{3}}{\mathrm{8}−\mathrm{3}}−\mathrm{2}<\mathrm{0} \\ $$$${k}=\mathrm{4} \\ $$$$\frac{\mathrm{2}+\mathrm{4}}{\mathrm{16}−\mathrm{3}}−\mathrm{2}<\mathrm{0} \\ $$$$ \\ $$
Answered by MJS last updated on 03/Apr/20
((∣2+log_2 x∣)/(x−3))<2 ⇒ x∈R\{0, 3} (1) 2+log_2 x ≥0 ⇒ x≥(1/4) ((2+log_2 x)/(x−3))<2 (1.1) (1/4)≤x<3 ⇒ x−3<0 256x>4^x ⇒ ≈.00392758<x<≈5.18752 ⇒ (1/4)≤x<3 • (1.2) x>3 ⇒ x−3>0 256x<4^x ⇒ x<≈.00392758∨x>≈5.28752 ⇒ x>≈5.28752 • (2) 2+log_2 x <0 ⇒ x<(1/4) ⇒ x−3<0 ((−(2+log_2 x))/(x−3))<2 always true because lhs<0 ⇒ x<0∨0<x<(1/4) • ⇒ x<0∨0<x<3∨x>≈5.28752
$$\frac{\mid\mathrm{2}+\mathrm{log}_{\mathrm{2}} \:{x}\mid}{{x}−\mathrm{3}}<\mathrm{2}\:\Rightarrow\:{x}\in\mathbb{R}\backslash\left\{\mathrm{0},\:\mathrm{3}\right\} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{2}+\mathrm{log}_{\mathrm{2}} \:{x}\:\geqslant\mathrm{0}\:\Rightarrow\:{x}\geqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{2}+\mathrm{log}_{\mathrm{2}} \:{x}}{{x}−\mathrm{3}}<\mathrm{2} \\ $$$$\left(\mathrm{1}.\mathrm{1}\right)\:\frac{\mathrm{1}}{\mathrm{4}}\leqslant{x}<\mathrm{3}\:\Rightarrow\:{x}−\mathrm{3}<\mathrm{0} \\ $$$$\mathrm{256}{x}>\mathrm{4}^{{x}} \:\Rightarrow\:\approx.\mathrm{00392758}<{x}<\approx\mathrm{5}.\mathrm{18752} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{4}}\leqslant{x}<\mathrm{3}\:\bullet \\ $$$$\left(\mathrm{1}.\mathrm{2}\right)\:{x}>\mathrm{3}\:\Rightarrow\:{x}−\mathrm{3}>\mathrm{0} \\ $$$$\mathrm{256}{x}<\mathrm{4}^{{x}} \:\Rightarrow\:{x}<\approx.\mathrm{00392758}\vee{x}>\approx\mathrm{5}.\mathrm{28752} \\ $$$$\Rightarrow\:{x}>\approx\mathrm{5}.\mathrm{28752}\:\bullet \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2}+\mathrm{log}_{\mathrm{2}} \:{x}\:<\mathrm{0}\:\Rightarrow\:{x}<\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:{x}−\mathrm{3}<\mathrm{0} \\ $$$$\frac{−\left(\mathrm{2}+\mathrm{log}_{\mathrm{2}} \:{x}\right)}{{x}−\mathrm{3}}<\mathrm{2}\:\mathrm{always}\:\mathrm{true}\:\mathrm{because}\:\mathrm{lhs}<\mathrm{0} \\ $$$$\Rightarrow\:{x}<\mathrm{0}\vee\mathrm{0}<{x}<\frac{\mathrm{1}}{\mathrm{4}}\:\bullet \\ $$$$ \\ $$$$\Rightarrow\:{x}<\mathrm{0}\vee\mathrm{0}<{x}<\mathrm{3}\vee{x}>\approx\mathrm{5}.\mathrm{28752} \\ $$
Commented by MJS last updated on 03/Apr/20
even if the question is wrong, we can solve it
$$\mathrm{even}\:\mathrm{if}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{wrong},\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{it} \\ $$
Commented by jagoll last updated on 03/Apr/20
but x < 0 not valid sir log_2 (x) ⇒ must be x > 0
$$\mathrm{but}\:\mathrm{x}\:<\:\mathrm{0}\:\mathrm{not}\:\mathrm{valid}\:\mathrm{sir} \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)\:\Rightarrow\:\mathrm{must}\:\mathrm{be}\:\mathrm{x}\:>\:\mathrm{0} \\ $$
Commented by MJS last updated on 03/Apr/20
you are right... but ∣2+log_2 x∣ ∈R∀x≠0 ... I′ll look into this again later
$$\mathrm{you}\:\mathrm{are}\:\mathrm{right}…\:\mathrm{but}\:\mid\mathrm{2}+\mathrm{log}_{\mathrm{2}} \:{x}\mid\:\in\mathbb{R}\forall{x}\neq\mathrm{0}\:… \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{look}\:\mathrm{into}\:\mathrm{this}\:\mathrm{again}\:\mathrm{later} \\ $$
Commented by MJS last updated on 03/Apr/20
log_2 x =((ln x)/(ln 2)) for x∈C ((ln x)/(ln 2))=((ln (re^(iθ) ))/(ln 2))=((ln r +iθ)/(ln 2)) ∣((ln r +iθ)/(ln 2))∣=((√(θ^2 +(ln r)^2 ))/(ln 2)) for x∈R^− : θ=π ⇒ ∣2+log_2 x∣≥0 ⇒ in our case for x<0 the equation is true
$$\mathrm{log}_{\mathrm{2}} \:{x}\:=\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\mathrm{for}\:{x}\in\mathbb{C} \\ $$$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{2}}=\frac{\mathrm{ln}\:\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)}{\mathrm{ln}\:\mathrm{2}}=\frac{\mathrm{ln}\:{r}\:+\mathrm{i}\theta}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\mid\frac{\mathrm{ln}\:{r}\:+\mathrm{i}\theta}{\mathrm{ln}\:\mathrm{2}}\mid=\frac{\sqrt{\theta^{\mathrm{2}} +\left(\mathrm{ln}\:{r}\right)^{\mathrm{2}} }}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\mathrm{for}\:{x}\in\mathbb{R}^{−} :\:\theta=\pi\:\Rightarrow\:\mid\mathrm{2}+\mathrm{log}_{\mathrm{2}} \:{x}\mid\geqslant\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:\mathrm{for}\:{x}<\mathrm{0}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{true} \\ $$
Commented by john santu last updated on 03/Apr/20
ooo if x ∈C sir. so the choice answer nothing correct sir
$$\mathrm{ooo}\:\mathrm{if}\:\mathrm{x}\:\in\mathbb{C}\:\mathrm{sir}.\: \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{choice}\:\mathrm{answer}\:\mathrm{nothing}\:\mathrm{correct} \\ $$$$\mathrm{sir} \\ $$

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