Find the value of $x$
$\frac{x^{\sqrt{x}}} {\sqrt{x}}=\frac{{\sqrt{x}}^x}{\sqrt{x}}$

Since both sides of the equation have a common denominator, multiply through by $\sqrt{x}$. 
$x^\sqrt{x}=\sqrt{x}^x$
$x^{x^\frac{1}{2}}=(x^\frac{1}{2})^x$
$x^{x^\frac{1}{2}}={x^{\frac{1}{2}}}^x$
🤔 Remember the axiom of indices with same bases on both sides of equation, applying that axiom yields, 
👉 $x^\frac{1}{2}=\frac{1}{2}x$
Take the squares of both sides 
$x=(\frac{1}{2})^2$
$x=\frac{1}{4}x^2$
This becomes a quadratic equations 
$4x=x^2$
Rearrange the equation and solve quadratically using any method you like. 
$x^2-4x=0$
Therefore, 
👉 $x=0,4$
And that brings us to the end of the solution. 



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