If $x^{x^x} = 2^{-\sqrt{2}}$, find $x^{-2}$

Question

I've been struggling with this one today. Applying the logarithmic function to both sides of the equation doesn't seem to work. Any ideas?

Answer 1

Because of $\enspace\displaystyle 2^\frac{1}{2}=(2^2)^\frac{1}{2^2}\enspace$ you can write

$$x^x\ln x = -\sqrt{2}\ln 2 = -\frac{2}{\sqrt{2}}\ln 2 = \left(\frac{1}{2}\right)^{\frac{1}{2}}\ln \frac{1}{2^2} = \left(\frac{1}{2^2}\right)^{\frac{1}{2^2}}\ln \frac{1}{2^2}$$

so that you get $\enspace\displaystyle x=\frac{1}{2^2}=0.25\enspace$ and therefore $\enspace\displaystyle x^{-2}=16$ .