They Are Inverted

Note that $x=0$ leads to an indeterminate form. So assume $x\neq 0$ .

$\begin{aligned} x^{x\sqrt{x}} &= (x\sqrt{x})^x \\ x^{x^{^3\!/\!_2}} &= x^{^3\!/\!_2x} \\ x^{^3\!/\!_2} &= \frac{3}{2}x && \color{#3D99F6}\text{Since }x\neq1 \\ x^{^1\!/\!_2} &= \frac{3}{2} \\ x &= \frac{9}{4} \end{aligned}$

Thus $a=9$ and $b=4$ , so $a+b=\boxed{13}$ .

$(x^{x \sqrt {x}})^{\frac{1}{x}} =((x\sqrt{x} )^{x})^{\frac{1}{x}}$

$x^{\sqrt{x}}=x\sqrt{x}$

$\frac {x^{\sqrt{x}}}{x}=\sqrt{x}$

$x^{\sqrt{x} -1}=\sqrt{x}$

square both sides.

$x^{2\sqrt{x} -2}=x$

$x^{2\sqrt{x} -3}=1$

$2\sqrt{x} -3=0$

$\sqrt{x} =\frac{3}{2}$

$x=\frac{9}{4}$