If $x^{x^4}=4$. Find $x^{x^2}+x^{x^8}$

Question

If $x^{x^4}=4$. Find $x^{x^2}+x^{x^8}$. I found this one in a competitive exam paper and found it interesting. Thanks for any help.

Answer 1

Let $4=x^{(x^2)}=\left((x^2)^{(x^2)}\right)^{x^2/2}$

$$x^2=y\implies y^{(y^2)}=16$$

One of the value of $y$ is $2$

$$x^{(x^2)}+x^{(x^8)}=x^2+x^{16}=2+(2)^8$$