Question Number 193793 by Mingma last updated on 20/Jun/23
Commented by MM42 last updated on 20/Jun/23
$${yes}.{you}\:{are}\:{right} \\ $$$${it}\:{was}\:{my}\:{mistake}.{thank}\:{you} \\ $$
Commented by Frix last updated on 20/Jun/23
$$\mathrm{1}.\mathrm{61053599}<{x}<\mathrm{1}.\mathrm{63195612} \\ $$
Commented by Frix last updated on 20/Jun/23
$$\lfloor\mathrm{3}.\mathrm{14}\rfloor=\mathrm{3}\:\mathrm{but}\:\lceil\mathrm{3}.\mathrm{14}\rceil=\mathrm{4} \\ $$$$ \\ $$$${x}=\mathrm{1}.\mathrm{62} \\ $$$$\lfloor\mathrm{1}.\mathrm{62}^{\lceil\mathrm{1}.\mathrm{62}^{\lceil\mathrm{1}.\mathrm{62}\rceil} \rceil\mathrm{1}.\mathrm{62}} \rfloor=\lfloor\mathrm{1}.\mathrm{62}^{\lceil\mathrm{1}.\mathrm{62}^{\mathrm{2}} \rceil\mathrm{1}.\mathrm{62}} \rfloor= \\ $$$$=\lfloor\mathrm{1}.\mathrm{62}^{\mathrm{3}×\mathrm{1}.\mathrm{62}} \rfloor=\lfloor\mathrm{10}.\mathrm{429}…\rfloor=\mathrm{10} \\ $$
Commented by Mingma last updated on 20/Jun/23
perfect ��
Answered by mr W last updated on 22/Jun/23
$${if}\:{x}\geqslant\mathrm{2}:\: \\ $$$$\lfloor{x}^{\lceil{x}^{\lceil{x}\rceil} \rceil{x}} \rfloor\geqslant\mathrm{2}^{\mathrm{2}^{\mathrm{2}} ×\mathrm{2}} =\mathrm{2}^{\mathrm{8}} =\mathrm{256}\:>\mathrm{10} \\ $$$${if}\:{x}=\mathrm{1}: \\ $$$$\lfloor{x}^{\lceil{x}^{\lceil{x}\rceil} \rceil{x}} \rfloor=\mathrm{1}<\mathrm{10} \\ $$$$\Rightarrow\mathrm{1}<{x}<\mathrm{2},\:{i}.{e}.\:\lceil{x}\rceil=\mathrm{2} \\ $$$${if}\:{x}\leqslant\sqrt{\mathrm{2}}: \\ $$$$\lfloor{x}^{\lceil{x}^{\lceil{x}\rceil} \rceil{x}} \rfloor\leqslant\lfloor\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}\sqrt{\mathrm{2}}} \rfloor=\mathrm{2}<\mathrm{10} \\ $$$$\Rightarrow{x}>\sqrt{\mathrm{2}} \\ $$$$\lceil{x}^{\lceil{x}\rceil} \rceil=\mathrm{3},\:{i}.{e}.\:\mathrm{2}<{x}^{\mathrm{2}} \leqslant\mathrm{3}\:\Rightarrow{x}\leqslant\sqrt{\mathrm{3}} \\ $$$$\lfloor{x}^{\lceil{x}^{\lceil{x}\rceil} \rceil{x}} \rfloor=\lfloor{x}^{\mathrm{3}{x}} \rfloor=\mathrm{10} \\ $$$$\mathrm{10}\leqslant{x}^{\mathrm{3}{x}} <\mathrm{11} \\ $$$$\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{3}}} \leqslant{x}^{{x}} <\mathrm{11}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{3}\:{W}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{3}}\right)}\leqslant{x}<\frac{\mathrm{ln}\:\mathrm{11}}{\mathrm{3}\:{W}\left(\frac{\mathrm{ln}\:\mathrm{11}}{\mathrm{3}}\right)}\:\checkmark \\ $$$$ \\ $$$${i}.{e}.\:\approx\mathrm{1}.\mathrm{61053599}\leqslant{x}<\approx\mathrm{1}.\mathrm{63195611} \\ $$
Commented by Frix last updated on 26/Jun/23
$$\mathrm{Great}! \\ $$