Question-154573

Question Number 154573 by liberty last updated on 19/Sep/21

Commented by ARUNG_Brandon_MBU last updated on 19/Sep/21

$$−\frac{\mathrm{1}}{\mathrm{6}}<{x}<\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

log_8 ((1/3)−x)∙log_(∣2x+(1/3)∣) ((1/3)−x)>log_2 (((1/3)−x)/( (((2x+(1/3))^2 ))^(1/3) )) (1/3)log_2 ((1/3)−x)∙((log_2 ((1/3)−x))/(log_2 (2x+(1/3))))>log_2 ((1/3)−x)−(2/3)log_2 (2x+(1/3)) (u^2 /(3v))>u−(2/3)v⇒u^2 >3uv−2v^2 ⇒(u−v)(u−2v)>0 log_2 (((1−3x)/(1+6x)))∙log_2 (((3(1−3x))/((1+6x)^2 )))>0 Case 1; ((1−3x)/(1+6x))>1 ∧ ((1−3x)/((1+6x)^2 ))>(1/3) ((−9x)/(6x+1))>0 ∧ ((3(1−3x)−(36x^2 +12x+1))/(3(6x+1)^2 ))>0 −(1/6)<x<0 ∧ 36x^2 +21x−2<0 Case 2 x<−(1/6)∪ x>0 ∧ 36x^2 +21x−2>0 ⇒0<x<(1/3) ∧36x^2 +21x−2>0

$$\mathrm{log}_{\mathrm{8}} \left(\frac{\mathrm{1}}{\mathrm{3}}−{x}\right)\centerdot\mathrm{log}_{\mid\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{3}}\mid} \left(\frac{\mathrm{1}}{\mathrm{3}}−{x}\right)>\mathrm{log}_{\mathrm{2}} \frac{\frac{\mathrm{1}}{\mathrm{3}}−{x}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}−{x}\right)\centerdot\frac{\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}−{x}\right)}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{3}}\right)}>\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}−{x}\right)−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{3}{v}}>{u}−\frac{\mathrm{2}}{\mathrm{3}}{v}\Rightarrow{u}^{\mathrm{2}} >\mathrm{3}{uv}−\mathrm{2}{v}^{\mathrm{2}} \Rightarrow\left({u}−{v}\right)\left({u}−\mathrm{2}{v}\right)>\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{3}{x}}{\mathrm{1}+\mathrm{6}{x}}\right)\centerdot\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{3}{x}\right)}{\left(\mathrm{1}+\mathrm{6}{x}\right)^{\mathrm{2}} }\right)>\mathrm{0} \\ $$$$\mathrm{Case}\:\mathrm{1}; \\ $$$$\frac{\mathrm{1}−\mathrm{3}{x}}{\mathrm{1}+\mathrm{6}{x}}>\mathrm{1}\:\wedge\:\frac{\mathrm{1}−\mathrm{3}{x}}{\left(\mathrm{1}+\mathrm{6}{x}\right)^{\mathrm{2}} }>\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{−\mathrm{9}{x}}{\mathrm{6}{x}+\mathrm{1}}>\mathrm{0}\:\wedge\:\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{3}{x}\right)−\left(\mathrm{36}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{6}{x}+\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}<{x}<\mathrm{0}\:\wedge\:\mathrm{36}{x}^{\mathrm{2}} +\mathrm{21}{x}−\mathrm{2}<\mathrm{0} \\ $$$$\mathrm{Case}\:\mathrm{2} \\ $$$${x}<−\frac{\mathrm{1}}{\mathrm{6}}\cup\:{x}>\mathrm{0}\:\wedge\:\mathrm{36}{x}^{\mathrm{2}} +\mathrm{21}{x}−\mathrm{2}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}<{x}<\frac{\mathrm{1}}{\mathrm{3}}\:\wedge\mathrm{36}{x}^{\mathrm{2}} +\mathrm{21}{x}−\mathrm{2}>\mathrm{0} \\ $$

Answered by iloveisrael last updated on 20/Sep/21

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