Just An Equation

First note that $x = 0$ is not a solution, since $0*\sqrt{0} = 0$ but $0^{\sqrt{0}} = 0^{0}$ is indeterminate. So given that $x \ne 0$ , we can write this equation as

$\large \dfrac{x^{\sqrt{x}}}{x^{\frac{3}{2}}} = 1 \Longrightarrow x^{(\sqrt{x} - \frac{3}{2})} = 1.$

Now since $1^{y} = 1$ for any real $y$ we have that $x = 1$ is a solution.

For $x \ne 1,0$ we must have that $\sqrt{x} - \dfrac{3}{2} = 0 \Longrightarrow x = \left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}.$

The sum of all possible solutions is thus $1 + \dfrac{9}{4} = \boxed{\dfrac{13}{4}}.$

$x^{\sqrt{x}}=x \sqrt{x}$ $\sqrt{x}\ln{x}=\ln{(x\sqrt{x})}$ $\sqrt{x}\ln{x}-\ln{x}-\ln{x^{1/2}}=0$ $\sqrt{x}\ln{x}-\ln{x}-\frac{1}{2}\ln{x}=0$ $(\sqrt{x}-1-\frac{1}{2})\ln{x}=0$ Hence, the solutions are: $\ln{x}=0 \Rightarrow x=1$ or: $\sqrt{x}-\frac{3}{2}=0 \Rightarrow x=9/4$ $1+\frac{9}{4}=\boxed{\frac{13}{4}}$

From observation, $1$ is a solution, and $0$ is not as ${0}^{0}$ is indeterminate. We also have:

$\ln { \left( { x }^{ \sqrt { x } } \right) } =\ln { \left( x\sqrt { x } \right) }$

$\sqrt { x } \ln { x } =\ln \left( {x}^{\large\frac{3}{2}} \right) =\dfrac{3}{2} \ln {x}$

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

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

Therefore the sum of the solutions is $1+\dfrac{9}{4}=\large \color{#20A900}{\boxed{\dfrac{13}{4}}}$ .

$x^{\sqrt{x}}=x\sqrt{x}$ $e^{\sqrt{x}lnx}=e^{ln(x\sqrt{x})}$ $\sqrt{x}lnx=\frac{3}{2} \cdot lnx$ $\rightarrow x_{1}=1$ $\sqrt{x}=\frac{3}{2}$ $\rightarrow x_{2}=\frac{4}{9}$ $x_{1}+x_{2}=\frac{13}{4}$