Climbing up the social ladder

Let $y = x^4$ . Then we have $x^{x^4} = x^{4\dot{}\frac{x^4}{4}} = y^{\frac{y}{4}} = 4$

$\implies y^y = 4^4 \implies y = x^4 = 4 \implies x = \left( 2^2 \right)^\frac{1}{4} = \sqrt{2}$

$\implies x^{x^2} + x^{x^8} = \sqrt{2}^{ \sqrt{2}^2} + \sqrt{2}^{ \sqrt{2}^8} = \sqrt{2}^{2} + \sqrt{2}^{16} = 2 + 2^8 = \boxed{258}$

In response to Moderator's note: We note that $\dfrac d{dy} y^y = (\ln y + 1)y^y > 0$ for $y > \frac 1e$ . Implying that $y^y$ is an increasing function for $y > \frac 1e$ . This means that $y^y = k$ has an unique solution when $k > \frac 1{e^\frac 1e}$ .

Simple,

Let $\large y = x^{x^{x^{x^{\cdots}}}}$

$\displaystyle \begin{aligned} x^y &= y\\x^{x^y}&=y\\ \text{Let } y&=4\\ \Rightarrow x^{x^4}&=4\\\Rightarrow x^4&=4\\\huge\Rightarrow x &= \huge 2^{\frac{1}{2}} \\x^{x^2}&=x^2\\&=\boxed{2}\\x^{x^8}&=x^{2^4}\\&=x^{16}\\&=2^8\\\Large&=\boxed{256}\end{aligned}$

$\Huge\therefore \boxed{x^{x^2}+x^{x^8} = 258}$

$x=a^{\frac{1}{2}} \\ x^{x^{4}} = a^{\frac{a^2}{2}} = (a^{a^2})^{\frac{1}{2}} = 4 \\ \Rightarrow a^{a^2} = 16$

It's easy to see a = 2. Substituting back in gives us the correct value for x, which is $\sqrt{2}$ .

$\large x^{x^4} =x^x*x^x*x^x*x^x =(\sqrt2)^4.~~~~~x=\sqrt2.......\\....Exp.=2+256=258$