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

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

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

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

Commented by ZiYangLee last updated on 13/Dec/20

$$★★$$

Commented by MJS_new last updated on 13/Dec/20

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

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

$$Therewasatypo.Ihaveedited$$

Commented by MJS_new last updated on 13/Dec/20

$$\mathrm{I}\mathrm{thought}\mathrm{so}$$

Commented by ZiYangLee last updated on 13/Dec/20

$$\mathrm{yaa}\mathrm{I}\mathrm{had}\mathrm{seen}\mathrm{it}\mathrm{haha}...$$

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