Question Number 99681 by Algoritm last updated on 22/Jun/20
Answered by floor(10²Eta[1]) last updated on 22/Jun/20
$$\mathrm{1}−\mathrm{2}{x}>\mathrm{0}\Rightarrow{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2}{x}\neq\mathrm{1}\Rightarrow{x}\neq\mathrm{0} \\ $$$${but}\:{we}\:{know}\:{that}\:{for}\:{all}\:{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}−\mathrm{2018}\:{will}\:{be}\:{negative},\:{so}\:{for}\:{that} \\ $$$${expression}\:{be}\:{positive} \\ $$$$\Rightarrow{log}_{\mathrm{1}−\mathrm{2}{x}} \left(\mathrm{2019}^{{x}} +\mathrm{2019}^{−{x}} \right)<\mathrm{0} \\ $$$$\Rightarrow\frac{{log}\left(\mathrm{2019}^{{x}} +\mathrm{2019}^{−{x}} \right)}{{log}\left(\mathrm{1}−\mathrm{2}{x}\right)}<\mathrm{0} \\ $$$${but}\:{log}\left(\mathrm{2019}^{{x}} +\mathrm{2019}^{−{x}} \right)>\mathrm{0}\:\forall\:{x}\in{R} \\ $$$${so},\:{log}\left(\mathrm{1}−\mathrm{2}{x}\right)<\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}{x}<\mathrm{1}\Rightarrow{x}>\mathrm{0}. \\ $$$$ \\ $$$${Solution}:\:{x}\in\left(\mathrm{0},\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$