Question

If $\sf x^{(x^4)} = 4,$ then find the value of $\sf x^{(x^2)} + x^{(x^8)}.$​

Answer 1

By inspection, we can see that x = ±√2, since (±√2)⁴ = 2² = 4.

In particular, if $x^n=n$, which means $x=n^{\frac1n}$, then $x^{x^n}=x^n=n$.

Let x = √2. Then

$x^{x^2}+x^{x^8}=(\sqrt2)^{(\sqrt2)^2}+(\sqrt2)^{(\sqrt2)^8}$

$x^{x^2}+x^{x^8}=(\sqrt2)^2+(\sqrt2)^{16}$

$x^{x^2}+x^{x^8}=2+256$

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