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Question Number 120927 by Ar Brandon last updated on 04/Nov/20

Answered by 675480065 last updated on 04/Nov/20

domain: 2x−3/4 >0 ∩ x>0 ∩ x>1 ⇒ x>3/8 ∩ x>0 hence 2x−3/4 > x^2 ⇒ x^2 −2x+3/4 <0 ⇒ (1/2)<x<(3/2) Final ans: ((3/8), (1/2))∪(1,(3/2)) B

$$\mathrm{domain}:\:\mathrm{2x}−\mathrm{3}/\mathrm{4}\:>\mathrm{0}\:\cap\:\mathrm{x}>\mathrm{0}\:\cap\:\mathrm{x}>\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{x}>\mathrm{3}/\mathrm{8}\:\cap\:\mathrm{x}>\mathrm{0} \\ $$$$\mathrm{hence}\:\mathrm{2x}−\mathrm{3}/\mathrm{4}\:>\:\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{3}/\mathrm{4}\:<\mathrm{0} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}<\mathrm{x}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{Final}\:\mathrm{ans}:\: \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{8}},\:\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left(\mathrm{1},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\mathrm{B} \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 04/Nov/20

Thanks Sir. But I got ((1/2),1)∪(1,(3/2)) Any opinion, please ?

$$\mathrm{Thanks}\:\mathrm{Sir}.\: \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{got}\:\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right)\cup\left(\mathrm{1},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\mathrm{Any}\:\mathrm{opinion},\:\mathrm{please}\:? \\ $$

Commented by Ar Brandon last updated on 04/Nov/20

Since for domaine we have 2x−3/4>0 ∧ x>0, x≠1 ⇒x>3/8 ∧ x>0, x≠1 and (1/2)<x<(3/2)

$$\mathrm{Since}\:\mathrm{for}\:\mathrm{domaine}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{2x}−\mathrm{3}/\mathrm{4}>\mathrm{0}\:\wedge\:\mathrm{x}>\mathrm{0},\:\mathrm{x}\neq\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}>\mathrm{3}/\mathrm{8}\:\wedge\:\mathrm{x}>\mathrm{0},\:\mathrm{x}\neq\mathrm{1} \\ $$$$\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{2}}<\mathrm{x}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by 675480065 last updated on 04/Nov/20

this is false

$$\mathrm{this}\:\mathrm{is}\:\mathrm{false} \\ $$$$ \\ $$

Answered by liberty last updated on 04/Nov/20

(1)⇔ log _x (2x−(3/4))>log _x (x^2 ) ⇔(x−1)(2x−(3/4)−x^2 )>0 ⇒(x−1)(4x^2 −8x+3)<0 ⇒(x−1)(2x−1)(2x−3)<0 ⇒ x<(1/2) ∪ 1<x<(3/2) (2)⇔numerus > 0 ; 2x>(3/4) ; x>(3/8) (3)⇔ base x>0 therefore we get solution from (1)∩(2)∩(3) ⇒ ((3/8),(1/2)) ∪ (1,(3/2))

$$\left(\mathrm{1}\right)\Leftrightarrow\:\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{4}}\right)>\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{x}^{\mathrm{2}} \right)>\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{4x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{3}\right)<\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{2x}−\mathrm{1}\right)\left(\mathrm{2x}−\mathrm{3}\right)<\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{x}<\frac{\mathrm{1}}{\mathrm{2}}\:\cup\:\mathrm{1}<\mathrm{x}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\mathrm{numerus}\:>\:\mathrm{0}\:;\:\mathrm{2x}>\frac{\mathrm{3}}{\mathrm{4}}\:;\:\mathrm{x}>\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\left(\mathrm{3}\right)\Leftrightarrow\:\mathrm{base}\:\mathrm{x}>\mathrm{0}\: \\ $$$$\mathrm{therefore}\:\mathrm{we}\:\mathrm{get}\:\mathrm{solution}\:\mathrm{from} \\ $$$$\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\cap\left(\mathrm{3}\right)\:\Rightarrow\:\left(\frac{\mathrm{3}}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{2}}\right)\:\cup\:\left(\mathrm{1},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$

Commented by Ar Brandon last updated on 04/Nov/20

Thanks for your efforts Sir

Answered by ebi last updated on 04/Nov/20

log_x (2x−(3/4))>2 ((log (2x−(3/4)))/(log x))>2 case (A) log x>0 ∴ x>1 case (B) 2x−(3/4)>0 ∴ x>(3/8) case (C) log (2x−(3/4))>log x^2 2x−(3/4)>x^2 x^2 −2x+(3/4)<0 (x−(1/2))(x−(3/2))<0 ∴ (1/2)<x<(3/2) solution (A) ∩ (B) ∩ (C) x=((3/8),(1/2))∪(1,(3/2))

$$ \\ $$$${log}_{{x}} \left(\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{4}}\right)>\mathrm{2} \\ $$$$\frac{{log}\:\left(\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{4}}\right)}{{log}\:{x}}>\mathrm{2} \\ $$$$ \\ $$$${case}\:\left({A}\right) \\ $$$${log}\:{x}>\mathrm{0} \\ $$$$\therefore\:{x}>\mathrm{1} \\ $$$$ \\ $$$${case}\:\left({B}\right) \\ $$$$\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{4}}>\mathrm{0} \\ $$$$\therefore\:{x}>\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$ \\ $$$${case}\:\left({C}\right) \\ $$$${log}\:\left(\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{4}}\right)>{log}\:{x}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{4}}>{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{3}}{\mathrm{4}}<\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)<\mathrm{0} \\ $$$$\therefore\:\frac{\mathrm{1}}{\mathrm{2}}<{x}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$${solution} \\ $$$$\left({A}\right)\:\cap\:\left({B}\right)\:\cap\:\left({C}\right) \\ $$$${x}=\left(\frac{\mathrm{3}}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left(\mathrm{1},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$

Commented by Ar Brandon last updated on 04/Nov/20

Thank you

Answered by Ar Brandon last updated on 04/Nov/20

log_x (2x−(3/4))>2=log_x x^2 Domaine: 2x−(3/4)>0 ∧ x>0, x≠1 ⇒ x>(3/8) ∧ x>0, x≠1 ⇒Domaine is x∈((3/8), 1)∪ (1, +∞) Case 1: when x∈((3/8), 1) log_x (2x−(3/4))>log_x x^2 ⇒2x−(3/4)<x^2 , since function is decreasing for x∈((3/8), 1) ie log_x a>log_x b ⇒ a<b ⇒4x^2 −8x+3>0 ⇒ (2x−1)(2x−3)>0 ⇒x∈(−∞, (1/2))∪((3/2), +∞) But from initial condition x∈((3/8), 1) Therefore x∈((3/8), (1/2)) Case 2: when x>1 log_x (2x−(3/4))>log_x x^2 2x−(3/4)>x^2 , since function is increasing for x>1. ie log_x a>log_x b ⇒ a>b ⇒4x^2 −8x+3<0 ⇒ (2x−1)(2x−3)<0 ⇒(1/2)<x<(3/2) But from initial condition x>1 Therefore x∈(1, (3/2)) Combining case 1 and case 2 we obtain x∈((3/8), (1/2))∪(1, (3/2))

$$\mathrm{log}_{\mathrm{x}} \left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{4}}\right)>\mathrm{2}=\mathrm{log}_{\mathrm{x}} \mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{Domaine}:\:\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{4}}>\mathrm{0}\:\wedge\:\mathrm{x}>\mathrm{0},\:\mathrm{x}\neq\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{x}>\frac{\mathrm{3}}{\mathrm{8}}\:\wedge\:\mathrm{x}>\mathrm{0},\:\mathrm{x}\neq\mathrm{1}\: \\ $$$$\Rightarrow\mathrm{Domaine}\:\mathrm{is}\:\mathrm{x}\in\left(\frac{\mathrm{3}}{\mathrm{8}},\:\mathrm{1}\right)\cup\:\left(\mathrm{1},\:+\infty\right) \\ $$$$\mathrm{Case}\:\mathrm{1}:\:\mathrm{when}\:\mathrm{x}\in\left(\frac{\mathrm{3}}{\mathrm{8}},\:\mathrm{1}\right) \\ $$$$\mathrm{log}_{\mathrm{x}} \left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{4}}\right)>\mathrm{log}_{\mathrm{x}} \mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{4}}<\mathrm{x}^{\mathrm{2}} ,\:\mathrm{since}\:\mathrm{function}\:\mathrm{is}\:\mathrm{decreasing}\:\mathrm{for} \\ $$$$\mathrm{x}\in\left(\frac{\mathrm{3}}{\mathrm{8}},\:\mathrm{1}\right)\:\mathrm{ie}\:\mathrm{log}_{\mathrm{x}} \mathrm{a}>\mathrm{log}_{\mathrm{x}} \mathrm{b}\:\Rightarrow\:\mathrm{a}<\mathrm{b} \\ $$$$\Rightarrow\mathrm{4x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{3}>\mathrm{0}\:\Rightarrow\:\left(\mathrm{2x}−\mathrm{1}\right)\left(\mathrm{2x}−\mathrm{3}\right)>\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}\in\left(−\infty,\:\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left(\frac{\mathrm{3}}{\mathrm{2}},\:+\infty\right) \\ $$$$\mathrm{But}\:\mathrm{from}\:\mathrm{initial}\:\mathrm{condition}\:\mathrm{x}\in\left(\frac{\mathrm{3}}{\mathrm{8}},\:\mathrm{1}\right) \\ $$$$\mathrm{Therefore}\:\mathrm{x}\in\left(\frac{\mathrm{3}}{\mathrm{8}},\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{Case}\:\mathrm{2}:\:\mathrm{when}\:\mathrm{x}>\mathrm{1} \\ $$$$\mathrm{log}_{\mathrm{x}} \left(\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{4}}\right)>\mathrm{log}_{\mathrm{x}} \mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{2x}−\frac{\mathrm{3}}{\mathrm{4}}>\mathrm{x}^{\mathrm{2}} ,\:\mathrm{since}\:\mathrm{function}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{for} \\ $$$$\mathrm{x}>\mathrm{1}.\:\mathrm{ie}\:\mathrm{log}_{\mathrm{x}} \mathrm{a}>\mathrm{log}_{\mathrm{x}} \mathrm{b}\:\Rightarrow\:\mathrm{a}>\mathrm{b} \\ $$$$\Rightarrow\mathrm{4x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{3}<\mathrm{0}\:\Rightarrow\:\left(\mathrm{2x}−\mathrm{1}\right)\left(\mathrm{2x}−\mathrm{3}\right)<\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}<\mathrm{x}<\frac{\mathrm{3}}{\mathrm{2}}\: \\ $$$$\mathrm{But}\:\mathrm{from}\:\mathrm{initial}\:\mathrm{condition}\:\mathrm{x}>\mathrm{1} \\ $$$$\mathrm{Therefore}\:\mathrm{x}\in\left(\mathrm{1},\:\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\mathrm{Combining}\:\mathrm{case}\:\mathrm{1}\:\mathrm{and}\:\mathrm{case}\:\mathrm{2}\:\mathrm{we}\:\mathrm{obtain} \\ $$$$\mathrm{x}\in\left(\frac{\mathrm{3}}{\mathrm{8}},\:\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left(\mathrm{1},\:\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$

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