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

Commented by ARUNG_Brandon_MBU last updated on 19/Sep/21

$$-\frac{1}{6}<x<\frac{1}{3}$$

Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

$${\log }_8(\frac{1}{3}-x)\cdot {\log }_{\mid 2x+\frac{1}{3}\mid }(\frac{1}{3}-x)>{\log }_2\frac{\frac{1}{3}-x}{\sqrt[3]{{(2x+\frac{1}{3})}^2}}$$$$\frac{1}{3}{\log }_2(\frac{1}{3}-x)\cdot \frac{{\log }_2(\frac{1}{3}-x)}{{\log }_2(2x+\frac{1}{3})}>{\log }_2(\frac{1}{3}-x)-\frac{2}{3}{\log }_2(2x+\frac{1}{3})$$$$\frac{u^2}{3v}>u-\frac{2}{3}v\Rightarrow u^2>3uv-2v^2\Rightarrow (u-v)(u-2v)>0$$$${\log }_2\left(\frac{1-3x}{1+6x}\right)\cdot {\log }_2\left(\frac{3(1-3x)}{{(1+6x)}^2}\right)>0$$$$\mathrm{Case}1;$$$$\frac{1-3x}{1+6x}>1\wedge \frac{1-3x}{{(1+6x)}^2}>\frac{1}{3}$$$$\frac{-9x}{6x+1}>0\wedge \frac{3(1-3x)-(36x^2+12x+1)}{3{(6x+1)}^2}>0$$$$-\frac{1}{6}<x<0\wedge 36x^2+21x-2<0$$$$\mathrm{Case}2$$$$x<-\frac{1}{6}\cup x>0\wedge 36x^2+21x-2>0$$$$\Rightarrow 0<x<\frac{1}{3}\wedge 36x^2+21x-2>0$$

Answered by iloveisrael last updated on 20/Sep/21

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