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Question Number 193626 by SaRahAli last updated on 17/Jun/23

Answered by MM42 last updated on 17/Jun/23

$$x=\sqrt{2}$$$$\Rightarrow {\left(\sqrt{2}\right)}^{{\left(\sqrt{2}\right)}^8}={\left(\sqrt{2}\right)}^{16}=256\mathrm{\&}{\left(\sqrt{2}\right)}^{{\left(\sqrt{2}\right)}^2}={\left(\sqrt{2}\right)}^2=2$$$$\Rightarrow and=258$$$$$$

Commented by aba last updated on 17/Jun/23

$${\left(\sqrt{2}\right)}^{16}=2^8=256$$

Commented by MM42 last updated on 17/Jun/23

$$yes.$$

Answered by aba last updated on 17/Jun/23

$${\mathrm{x}}^{{\mathrm{x}}^4}=4\Rightarrow {\mathrm{x}}^4\mathrm{lnx}=2\ln 2\Rightarrow 4\mathrm{lnx}.{\mathrm{e}}^{4\mathrm{lnx}}=8\ln 2$$$$\Rightarrow 4\mathrm{lnx}=\mathrm{W}\left(8\ln 2\right)=\mathrm{W}(2\ln 2.{\mathrm{e}}^{2\ln 2})=2\ln 2$$$$\Rightarrow \mathrm{lnx}=\frac{1}{2}\ln 2\Rightarrow \mathrm{x}=\sqrt{2}$$$$$$$${\left(\sqrt{2}\right)}^{{\left(\sqrt{2}\right)}^8}+{\left(\sqrt{2}\right)}^{{\left(\sqrt{2}\right)}^2}={\left(\sqrt{2}\right)}^{{\left(\sqrt{2}\right)}^{2\times 4}}+{\left(\sqrt{2}\right)}^{{\left(\sqrt{2}\right)}^2}={\left(\sqrt{2}\right)}^{2^4}+{\left(\sqrt{2}\right)}^{{\left(\sqrt{2}\right)}^2}$$$$={\left(\sqrt{2}\right)}^{2\times 8}+{\left(\sqrt{2}\right)}^2$$$$=2^8+2$$$$=258✓$$$$$$

Answered by mr W last updated on 17/Jun/23

$$x^{x^4}=4$$$${\left(x^4\right)}^{x^4}=4^4$$$$\Rightarrow x^4=4$$$$\Rightarrow x^2=2$$$$x^{x^8}=x^{{\left(x^4\right)}^2}=x^{16}={\left(x^4\right)}^4=4^4=256$$$$x^{x^2}=x^2=2$$$$\Rightarrow x^{x^8}+x^{x^2}=256+2=258$$

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