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Question Number 84674 by jagoll last updated on 15/Mar/20

$$\frac{6-{\log }_{16}\left({\mathrm{x}}^4\right)}{3+{2\log }_{16}\left({\mathrm{x}}^2\right)}<2$$

Commented byjagoll last updated on 15/Mar/20

$$\left(\mathrm{i}\right)\mathrm{x}\ne 0$$ $$\Rightarrow \frac{6-{2\log }_{16}\left({\mathrm{x}}^2\right)}{3+{2\log }_{16}\left({\mathrm{x}}^2\right)}<2$$ $$\Rightarrow \frac{3-{\log }_{16}\left({\mathrm{x}}^2\right)}{3+{2\log }_{16}\left({\mathrm{x}}^2\right)}<1$$ $$\mathrm{let}\mathrm{t}={\log }_{16}\left({\mathrm{x}}^2\right)$$ $$\frac{3-\mathrm{t}}{3+2\mathrm{t}}-\frac{3+2\mathrm{t}}{3+2\mathrm{t}}<0$$ $$\frac{-3\mathrm{t}}{3+2\mathrm{t}}<0\Rightarrow \mathrm{t}<-\frac{3}{2}\vee \mathrm{t}>0$$ $${\log }_{16}\left({\mathrm{x}}^2\right)<{\log }_{16}{\left(16\right)}^{-\frac{3}{2}}\vee$$ $${\log }_{16}\left({\mathrm{x}}^2\right)>{\log }_{16}{\left(16\right)}^0$$ $${\mathrm{x}}^2<\frac{1}{64}\vee {\mathrm{x}}^2>1\Rightarrow \mathrm{x}<-1\vee$$ $$-\frac{1}{8}<\mathrm{x}<\frac{1}{8}\vee \mathrm{x}>1$$ $$$$ $$$$

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