Menu Close

Question-191966




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^(yln x)   (−yln x)e^(−yln x) =−ln x  −y ln x=W(−ln x)  y=−((W(−ln x))/(ln x))  −ln x≥−(1/e)  ln x≤(1/e)  ⇒0<x≤(e)^(1/e)  ≈1.444668
$${y}={x}^{{x}^{…} } ={x}^{{y}} \\ $$$${y}={e}^{{y}\mathrm{ln}\:{x}} \\ $$$$\left(−{y}\mathrm{ln}\:{x}\right){e}^{−{y}\mathrm{ln}\:{x}} =−\mathrm{ln}\:{x} \\ $$$$−{y}\:\mathrm{ln}\:{x}=\mathbb{W}\left(−\mathrm{ln}\:{x}\right) \\ $$$${y}=−\frac{\mathbb{W}\left(−\mathrm{ln}\:{x}\right)}{\mathrm{ln}\:{x}} \\ $$$$−\mathrm{ln}\:{x}\geqslant−\frac{\mathrm{1}}{{e}} \\ $$$$\mathrm{ln}\:{x}\leqslant\frac{\mathrm{1}}{{e}} \\ $$$$\Rightarrow\mathrm{0}<{x}\leqslant\sqrt[{{e}}]{{e}}\:\approx\mathrm{1}.\mathrm{444668} \\ $$
Commented by Shlock last updated on 04/May/23
Nice solution, sir!

Leave a Reply

Your email address will not be published. Required fields are marked *