If $x^{x^4} = 4$, what is the value of $x^{x^2} + x^{x^8}$? If $x^{x^4} = 4$, what is the value of $x^{x^2} + x^{x^8}$ ?
I can find by trial and error, that $x=\sqrt 2$. But, what is the general process to answer questions like this?

Taking the logarithm base 2 of each side, we have
$\log_2 (x^{x^4}) = x^4\log_2 (x)= \log_2 4 = 2$
Multiplying each side by four we have
$4x^4\log_2 (x) = x^4\log_2 (x^4) = 8$
Relabeling $x^4$ as $u$, we have
$u\log_2(u) = 8$
Utilizing the Lambert W Function, we have that
$u = e^{W(8\ln 2)}=e^{W(4\ln 4)}=4$
Replacing back, $x^4=4$ and the result follows.

For the conclusion to the problem, we found earlier that $x^4=4$ and so either $x^2=2$ or $x^2=-2$.  Assuming that we require $x$ to be real, it must be the first.
$\begin{array}{rl}
x^{x^2} + x^{x^8} &=x^2 + x^{(x^4)^2}\\
&=2+x^{16}\\
&=2+(x^4)^4\\
&=2+4^4\\
&=2+256\\
&=258\end{array}$
