How to get the value of $x$? I have this statement:

Get the value of $x$ given that $x^{\sqrt{x}}= \sqrt[\frac{1}{\sqrt{2}}]{\dfrac{1}{\sqrt{2}}}$ .

My attempt was not very useful, but this is:
Let $a = \dfrac {1}{\sqrt{2}}$, then $x^\sqrt{x}= a^{\frac{1}{a}}$
And I can't get more. Any hint is appreciated.
 A: If $a = \frac 1{\sqrt 2} = 2^{- \frac 12}$, then $x^{\sqrt x} = a^{\frac 1a} = 2^{\frac {-1}{2a}}$.
Let $x = 2^y$. Then, $x^{\sqrt x} = 2^{y2^{\frac y2}}$.
Then, $y2^{\frac y2} = \frac{-1}{2a} = -\frac 1{\sqrt 2}$. Clearly $y < 0$, let $z = -y$. Taking $2$ powers gives the equation $\frac {-1}2 = \ln z - \frac z2$. An easy check gives that $z = 1$ satisfies this (no further simplification can be done at this point other than checking) and therefore $x = \frac 12$. There is one more real solution of $z$, but this can only be found numerically. 
A: Let's try to rewrite RHS a bit to make it of the form on LHS:
We have $$\frac{1}{\sqrt 2^{\sqrt 2}}=\left( 2^{-1/2}\right)^{\sqrt 2}=2^{-\sqrt 2/2}=2^{-1/\sqrt 2},$$ or to make the LHS form explicit, $$\left(\frac{1}{2}\right)^{\sqrt{\frac{1}{2}}},$$ whence $x=1/2.$
A: Using your hint, we draw
$$\sqrt x^{\sqrt x}=a^{1/2a}.$$
Given the particular value of $a$, we have
$$\frac1{2a}=a,$$ so that 
$$\sqrt x^{\sqrt x}=a^a$$ yields $$x=a^2=\frac12.$$

Given that $t^t$ has a minimum at $(e^{-1},e^{-e^{-1}})$ and that $0^0=1$, there is another solution for $\sqrt x$ in $(0,e^{-1})$.
A: This may be similar to other solutions. 
$$\begin{align}
x^{\sqrt{x}}&=\sqrt[\frac 1{\sqrt2}]{\frac 1{\sqrt{2}}}\\
&=\left(\frac 1{\sqrt{2}}\right)^{\sqrt{2}}\\
&=\frac 1{\sqrt{2}^\sqrt{2}}\\
&=\frac 1{(2^{1/2})^\sqrt{2}}\\
&=\frac 1{2^{1/\sqrt{2}}}\\
&=\left(\frac 12\right)^{\frac 1{\sqrt{2}}}\\
x&=\frac 12
\end{align}
$$
