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Question Number 81364 by jagoll last updated on 12/Feb/20

$${\log }_2({\log }_3\left({\log }_2\frac{2}{\mathrm{x}}\right)-1)<1$$ $$\mathrm{has}\mathrm{solution}\frac{1}{{\mathrm{a}}^{26}}<\mathrm{x}<\mathrm{b}.\mathrm{find}\mathrm{a}?$$

Commented byjohn santu last updated on 12/Feb/20

$$\left(\mathrm{i}\right){\log }_3\left({\log }_2\frac{2}{\mathrm{x}}\right)-1>0$$ $$\left(\mathrm{ii}\right){\log }_3\left({\log }_2\frac{2}{\mathrm{x}}\right)-1<2$$ $$\Rightarrow 0<{\log }_3\left({\log }_2\frac{2}{\mathrm{x}}\right)-1<2$$ $$\Rightarrow 1<{\log }_3\left({\log }_2\frac{2}{\mathrm{x}}\right)<3$$ $$\Rightarrow 3<{\log }_2\left(\frac{2}{\mathrm{x}}\right)<27$$ $$\Rightarrow 2^3<\frac{2}{\mathrm{x}}<2^{27}$$ $$\Rightarrow \frac{1}{2^{27}}<\frac{\mathrm{x}}{2}<\frac{1}{2^3}$$ $$\Rightarrow \frac{1}{2^{26}}<\mathrm{x}<\frac{1}{2^2}.\mathrm{we}\mathrm{get}\mathrm{a}=2$$

Commented byjagoll last updated on 12/Feb/20

$$\mathrm{thank}\mathrm{mister}$$

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