Potenciação com incógnitas e raiz

[petras]

\(x^{x^x}=2^{-\sqrt{2}}\)

Então \(x^{-2}\)

[petras]

Alternativas:

A) 32
B) 16
C) 8
D) 4
E) 1/2

[petras]

Compartilho a resolução do colega "vanstraelen"
\(x^{x^x}= 2^{-\sqrt2}=(4^{\frac{1}{2}})^{-\sqrt2}=4^{-\frac{\sqrt{2}}{2}}=(4^{-1})^{\frac{\sqrt{2}}{2}}=(4^{-1})^{2^{-\frac{1}{2}}}=(4^{-1})^{4^{-\frac{1}{4}}}=[4^{-1}]^(4^{-1})^{\frac{1}{4}}=\frac{1}{4}^{\frac{1}{4}^{\frac{1}{4}}}\rightarrow x= \frac{1}{4}\rightarrow \frac{1}{4}^{-2}=4^2 = 16\)