Question Number 191966 by Shlock last updated on 04/May/23
Answered by mr W last updated on 04/May/23
$${y}={x}^{{x}^{...} } ={x}^{{y}} \\ $$$${y}={e}^{{y}\mathrm{ln}\:{x}} \\ $$$$\left(−{y}\mathrm{ln}\:{x}\right){e}^{−{y}\mathrm{ln}\:{x}} =−\mathrm{ln}\:{x} \\ $$$$−{y}\:\mathrm{ln}\:{x}=\mathbb{W}\left(−\mathrm{ln}\:{x}\right) \\ $$$${y}=−\frac{\mathbb{W}\left(−\mathrm{ln}\:{x}\right)}{\mathrm{ln}\:{x}} \\ $$$$−\mathrm{ln}\:{x}\geqslant−\frac{\mathrm{1}}{{e}} \\ $$$$\mathrm{ln}\:{x}\leqslant\frac{\mathrm{1}}{{e}} \\ $$$$\Rightarrow\mathrm{0}<{x}\leqslant\sqrt[{{e}}]{{e}}\:\approx\mathrm{1}.\mathrm{444668} \\ $$
Commented by Shlock last updated on 04/May/23
Nice solution, sir!