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Question Number 124192 by Algoritm last updated on 01/Dec/20

Commented by MJS_new last updated on 01/Dec/20

$$2$$

Commented by Algoritm last updated on 01/Dec/20

$$\mathrm{proof}$$

Answered by MJS_new last updated on 01/Dec/20

$$2<\sqrt{3}{}^{\sqrt{3}}<3$$$$2<3^{\frac{\sqrt{3}}{2}}<3$$$$4<3^{\sqrt{3}}<9$$$$4^{\sqrt{3}}<27<9^{\sqrt{3}}$$$$\frac{3}{2}<\sqrt{3}<2\Rightarrow \{{}_{9^{\frac{3}{2}}<9^{\sqrt{3}}<9^2\iff 27<9^{\sqrt{3}}<81\Rightarrow 27<9^{\sqrt{3}}}^{4^{\frac{3}{2}}<4^{\sqrt{3}}<4^2\iff 8<4^{\sqrt{3}}<16\Rightarrow 4^{\sqrt{3}}<27}$$$$\Rightarrow 2<\sqrt{3}{}^{\sqrt{3}}<3\Rightarrow \lfloor \sqrt{3}{}^{\sqrt{3}}\rfloor =2$$

Commented by MJS_new last updated on 01/Dec/20

$$\mathrm{same}\mathrm{procedure}$$$$6<3^{\sqrt{3}}<7$$$$6^{\sqrt{3}}<27<7^{\sqrt{3}}$$$$\frac{17}{10}<\sqrt{3}<\frac{18}{10}\Rightarrow \{{}_{7^{\frac{17}{10}}<7^{\sqrt{3}}<7^{\frac{18}{10}}}^{6^{\frac{17}{10}}<6^{\sqrt{3}}<6^{\frac{18}{10}}}$$$$6^{\frac{18}{10}}<27\iff 6^{18}<{27}^{10}\iff 2^{18}{3}^{18}<3^{12}{3}^{18}\iff$$$$\iff 2^{18}<3^{12}\iff 2\sqrt{2}<3\mathrm{true}$$$$27<7^{\frac{17}{10}}\iff {27}^{10}<7^{17}\iff 205891132094649<232630513987207$$$$\mathrm{true}\left(\mathrm{sorry}\mathrm{I}\mathrm{found}\mathrm{no}\mathrm{better}\mathrm{way}\right)$$$$\Rightarrow 6^{\sqrt{3}}<6^{\frac{18}{10}}<27<7^{\frac{17}{10}}<7^{\sqrt{3}}$$$$\iff 6<3^{\sqrt{3}}<7$$

Commented by Algoritm last updated on 01/Dec/20

$$\mathrm{Brilliant}$$

Commented by Algoritm last updated on 01/Dec/20

Commented by MJS_new last updated on 01/Dec/20

$$\mathrm{what}\mathrm{do}\mathrm{you}\mathrm{mean}?$$$$3^{\sqrt{3}}\approx 6.70499$$

Commented by Algoritm last updated on 01/Dec/20

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