Question Number 215350 by hardmath last updated on 03/Jan/25
$$\mathrm{Find}:\:\:\:\mathrm{x}^{\boldsymbol{\mathrm{x}}} \:=\:\mathrm{2}^{\sqrt{\mathrm{200}}} \:\:\Rightarrow\:\:\mathrm{x}\:=\:? \\ $$
Answered by zetamaths last updated on 03/Jan/25
$$.{is}\:{just}\:{the}\:{lambert}\:{function} \\ $$$${x}^{{x}} =\mathrm{2}^{\sqrt{\mathrm{200}}} \\ $$$$=>\:\:{x}=\mathrm{2}^{\frac{\sqrt{\mathrm{200}}}{{x}}} \\ $$$$ \\ $$$${x}=\left({e}^{{ln}\mathrm{2}} \right)^{\frac{\sqrt{\mathrm{200}}}{{x}}} \\ $$$$\frac{\sqrt{\mathrm{200}}}{{x}}{ln}\mathrm{2}×{x}=\frac{\sqrt{\mathrm{200}}}{{x}}{ln}\mathrm{2}×\left({e}^{\frac{{ln}\mathrm{2}\sqrt{\mathrm{200}}}{{x}}} \right) \\ $$$${ln}\mathrm{2}×\sqrt{\mathrm{200}}={ve}^{{v}} \\ $$$${W}\left({ln}\mathrm{2}×\sqrt{\mathrm{200}}\right)={v} \\ $$$${x}=\frac{{ln}\mathrm{2}×\sqrt{\mathrm{200}}}{{W}\left({ln}\mathrm{2}×\sqrt{\left.\mathrm{200}\right)}\right.} \\ $$$${because}\:\:{v}={ln}\mathrm{2}\frac{\sqrt{\mathrm{200}}}{{x}} \\ $$$${and}\:{lambert}\:{function}\:{isW}\left({xe}^{{x}} \right)={x} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 03/Jan/25
![2^(√(200)) =2^(10(√2)) =((√2))^(20(√2)) =[((√2))^5 ]^(4(√2)) =(4(√2))^(4(√2)) x^x =2^(√(200)) =(4(√2))^(4(√2)) ⇒x=4(√2) ✓](https://www.tinkutara.com/question/Q215353.png)
$$\mathrm{2}^{\sqrt{\mathrm{200}}} =\mathrm{2}^{\mathrm{10}\sqrt{\mathrm{2}}} =\left(\sqrt{\mathrm{2}}\right)^{\mathrm{20}\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:=\left[\left(\sqrt{\mathrm{2}}\right)^{\mathrm{5}} \right]^{\mathrm{4}\sqrt{\mathrm{2}}} =\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$${x}^{{x}} =\mathrm{2}^{\sqrt{\mathrm{200}}} =\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{x}=\mathrm{4}\sqrt{\mathrm{2}}\:\:\checkmark \\ $$
Commented by hardmath last updated on 04/Jan/25
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$