Question Number 193626 by SaRahAli last updated on 17/Jun/23
Answered by MM42 last updated on 17/Jun/23
$${x}=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} } =\left(\sqrt{\mathrm{2}}\right)^{\mathrm{16}} =\mathrm{256}\:\:\&\:\:\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } =\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{and}=\mathrm{258} \\ $$$$ \\ $$
Commented by aba last updated on 17/Jun/23
$$\left(\sqrt{\mathrm{2}}\right)^{\mathrm{16}} =\mathrm{2}^{\mathrm{8}} =\mathrm{256} \\ $$
Commented by MM42 last updated on 17/Jun/23
$${yes}. \\ $$
Answered by aba last updated on 17/Jun/23
$$\mathrm{x}^{\mathrm{x}^{\mathrm{4}} } =\mathrm{4}\:\Rightarrow\:\mathrm{x}^{\mathrm{4}} \mathrm{lnx}=\mathrm{2ln2}\:\Rightarrow\:\mathrm{4lnx}.\mathrm{e}^{\mathrm{4lnx}} =\mathrm{8ln2} \\ $$$$\Rightarrow\:\mathrm{4lnx}=\mathrm{W}\left(\mathrm{8ln2}\right)=\mathrm{W}\left(\mathrm{2ln2}.\mathrm{e}^{\mathrm{2ln2}} \right)=\mathrm{2ln2} \\ $$$$\Rightarrow\:\mathrm{lnx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln2}\:\Rightarrow\:\mathrm{x}=\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} } +\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } =\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}×\mathrm{4}} } +\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } =\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}^{\mathrm{4}} } +\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}×\mathrm{8}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{8}} +\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{258}\:\checkmark \\ $$$$ \\ $$
Answered by mr W last updated on 17/Jun/23
$${x}^{{x}^{\mathrm{4}} } =\mathrm{4} \\ $$$$\left({x}^{\mathrm{4}} \right)^{{x}^{\mathrm{4}} } =\mathrm{4}^{\mathrm{4}} \\ $$$$\Rightarrow{x}^{\mathrm{4}} =\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{{x}^{\mathrm{8}} } ={x}^{\left({x}^{\mathrm{4}} \right)^{\mathrm{2}} } ={x}^{\mathrm{16}} =\left({x}^{\mathrm{4}} \right)^{\mathrm{4}} =\mathrm{4}^{\mathrm{4}} =\mathrm{256} \\ $$$${x}^{{x}^{\mathrm{2}} } ={x}^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{x}^{{x}^{\mathrm{8}} } +{x}^{{x}^{\mathrm{2}} } =\mathrm{256}+\mathrm{2}=\mathrm{258} \\ $$