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Question Number 99681 by Algoritm last updated on 22/Jun/20

Answered by floor(10²Eta[1]) last updated on 22/Jun/20

$$1-2x>0\Rightarrow x<\frac{1}{2}$$$$1-2x\ne 1\Rightarrow x\ne 0$$$$butweknowthatforallx<\frac{1}{2}$$$$x-2018willbenegative,soforthat$$$$expressionbepositive$$$$\Rightarrow {log}_{1-2x}({2019}^x+{2019}^{-x})<0$$$$\Rightarrow \frac{log({2019}^x+{2019}^{-x})}{log(1-2x)}<0$$$$butlog({2019}^x+{2019}^{-x})>0\mathrm{\forall }x\in R$$$$so,log(1-2x)<0$$$$\Rightarrow 1-2x<1\Rightarrow x>0.$$$$$$$$Solution:x\in (0,\frac{1}{2})$$

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