If $x^{x^4} = 4$, what is the value of $x^{x^2} + x^{x^8}$?

Question

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?

Answer 1

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.

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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.

$\begin{array}{rl} x^{x^2} + x^{x^8} &=x^2 + x^{(x^4)^2}\\ &=2+x^{16}\\ &=2+(x^4)^4\\ &=2+4^4\\ &=2+256\\ &=258\end{array}$