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

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

I tried to simplify it using exponentiation and logs, and even just algebraic manipulation..But I don't know how to do this.

• Linear algebra? To clarify, the expression is $x^{(x^4)}$ and so on? If so, then note that the real roots of $x^{(x^4)}=4$ are $±\sqrt 2$. – lulu Oct 6 '15 at 10:35
• Isn't the domain of $x^{x^4}$ should be $x>0$? – Quang Hoang Oct 6 '15 at 10:46
• Yes. @lulu That's what the question means. – Kugelblitz Oct 6 '15 at 10:50
• Note that if a number $x$ satisfies $x^4=4$, then it also satisfies $x^{x^4}=4$ – nospoon Oct 6 '15 at 10:54
• A cute trick for this kind of thing: if $x^{x^4}=4$ then $(x^4)^{x^4}=x^{4x^4}=(x^{x^4})^4=4^4$. This makes it easy to spot the solution $x^4=4$. (similar) – user21467 Oct 6 '15 at 11:52

Just note that $x^{x^4}$ is increasing for $x>1$ and is $<1$ for $x<1$. The unique positive solution to $x^{x^4}=4$ is $x=\sqrt2$.
$x^{x^{4}}=4$. Let us replace 4 of the exponent of left hand side by $x^{x^{4}}$ then equation becomes $x^{x^{x^{x^{4}}}}=4$. Repeating this process infinitely we get $x^{x^{x^{x^{x..}}}}=4$. Now We replace the exponent of left hand side by 4 and equation now become $x^{4}=4$ hence $x=\sqrt{2}$ or $x=-\sqrt{2}$ . So $x^{x^{2}}+x^{x^{8}} =258$