Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

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))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com