Given-that-x-x-4-4-find-the-value-of-x-x-2-x-x-8-

Question Number 125739 by ZiYangLee last updated on 13/Dec/20

$$\mathrm{Given}\:\mathrm{that}\:{x}^{{x}^{\mathrm{4}} } =\mathrm{4},\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}^{{x}^{\mathrm{2}} } +{x}^{{x}^{\mathrm{8}} } . \\ $$

Commented by Dwaipayan Shikari last updated on 13/Dec/20

x^x^4 =4 ⇒x^x^x^x^4 =4 ⇒ x^4 =4⇒x=±(√2) x^x^2 +x^x^8 =(±(√2))^2 +(±(√2))^(((√2))^8 ) =2+(±(√2))^(16) =258

$${x}^{{x}^{\mathrm{4}} } =\mathrm{4}\:\Rightarrow{x}^{{x}^{{x}^{{x}^{\mathrm{4}} } } } =\mathrm{4}\:\Rightarrow\:{x}^{\mathrm{4}} =\mathrm{4}\Rightarrow{x}=\pm\sqrt{\mathrm{2}} \\ $$$${x}^{{x}^{\mathrm{2}} } +{x}^{{x}^{\mathrm{8}} } =\left(\pm\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\pm\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} } =\mathrm{2}+\left(\pm\sqrt{\mathrm{2}}\right)^{\mathrm{16}} =\mathrm{258} \\ $$

Commented by ZiYangLee last updated on 13/Dec/20

$$\bigstar\bigstar \\ $$

Commented by MJS_new last updated on 13/Dec/20

x^x^a =x^((x^a )) (√2)^(((√2)^2 )) =(√2)^2 =2 (√2)^(((√2)^8 )) =(√2)^(16) =256 2+256=258

$${x}^{{x}^{{a}} } ={x}^{\left({x}^{{a}} \right)} \\ $$$$\sqrt{\mathrm{2}}\:^{\left(\sqrt{\mathrm{2}}\:^{\mathrm{2}} \right)} =\sqrt{\mathrm{2}}\:^{\mathrm{2}} =\mathrm{2} \\ $$$$\sqrt{\mathrm{2}}\:^{\left(\sqrt{\mathrm{2}}\:^{\mathrm{8}} \right)} =\sqrt{\mathrm{2}}\:^{\mathrm{16}} =\mathrm{256} \\ $$$$\mathrm{2}+\mathrm{256}=\mathrm{258} \\ $$

Commented by Dwaipayan Shikari last updated on 13/Dec/20