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Question Number 83774 by jagoll last updated on 06/Mar/20

$$\mathrm{Given}2\sqrt{{\log }_{3}x-1}-{\log }_3{x}^3+8>0$$ $$have\mathrm{the}\mathrm{solution}\mathrm{a}⩽x<b.$$ $$whatisb?$$

Answered by john santu last updated on 06/Mar/20

$$\Rightarrow 2\sqrt{{\log }_{3}x-1}-{3\log }_{3}x+8>0$$ $$\mathrm{let}\sqrt{{\log }_{3}x-1}=\mathrm{t},\mathrm{t}⩾0\left(\mathrm{i}\right)$$ $$\Rightarrow 2\mathrm{t}-3({\mathrm{t}}^2+1)+8>0$$ $${3\mathrm{t}}^2+3-2\mathrm{t}-8<0$$ $${3\mathrm{t}}^2-2\mathrm{t}-5<0$$ $$(3\mathrm{t}-5)(\mathrm{t}+1)<0\Rightarrow -1<\mathrm{t}<\frac{5}{3}\left(\mathrm{ii}\right)$$ $$\mathrm{from}\left(\mathrm{i}\right)\mathrm{\&}\left(\mathrm{ii}\right)$$ $$0⩽\mathrm{t}<\frac{5}{3}\Rightarrow 0⩽\sqrt{{\log }_{3}x-1}<\frac{5}{3}$$ $$0⩽{\log }_{3}x-1<\frac{25}{9}$$ $$1⩽{\log }_{3}x<\frac{34}{9}\iff 3^1⩽x<3^{\frac{34}{9}}$$ $$\mathrm{so}\mathrm{we}\mathrm{get}\mathrm{b}=3^{\frac{34}{9}}\Leftarrow \mathrm{the}\mathrm{solution}$$ $$$$

Commented byjagoll last updated on 06/Mar/20

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

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