Is this question even solvable? I am trying to solve $x^{x^{3/2}} = 3x/2$.  My teacher said that it's a closed equation and could not be solved.  Then I plotted this equation on Wolfram Alpha and the graph showed 2 solutions.  I tried it by solving it by logarithm but it became very complex.  Then I tried solving it by inequalities and the solution set was coming greater than 1, but on Wolfram Alpha the solutions are about 0.52 and 1.57.  
Is it even solvable?
 A: This is not solvable exactly, but the solution can be approximated via iteration.
We find $g : f(x)=0\iff g(x)=x$, then $$f(\lim_{n\to\infty} x_n)=0 \text{ where }x_{n+1}=g(x_n)$$
So we have: $x^{x\sqrt x}=\frac{3x}{2}\to x=\frac23x^{x\sqrt x} \ [1] \text{ and } x=\log_x^{\frac23}(\frac{3x}{2}) \ [2]$
Starting with $x_0=1$ for $[1]$ and $x_0=2$ for $[2]$ we get that 
$$[1] \to 0.5217085869$$
$$[2]\to 1.548591331$$
 both to calculator accuracy.
A: Since you graphed the function
$$f(x)=x^{x^{3/2}}-\frac{3 }{2}x$$ you noticed that it goes through a minimum "around" $x=1$ (which is a very nice value !).
Using Taylor series around $x=1$, we have
$$f(x)=-\frac{1}{2}-\frac{1}{2}(x-1)+\frac{3}{2} (x-1)^2+O\left((x-1)^3\right)$$ Neglecting the higher order terms and solving the quadratic equation in $(x-1)$ gives as estimates of the roots
$$x_1=\frac{7-\sqrt{13}}{6}\approx 0.565741\qquad \text{and} \qquad x_2=\frac{7+\sqrt{13}}{6}\approx 1.76759$$ Using these estimates as starting points for Newton method, we should get the dollowing iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0.5657414541 \\
 1 & 0.5205808928 \\
 2 & 0.5217078869
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 1.767591879 \\
 1 & 1.620506201 \\
 2 & 1.558309746 \\
 3 & 1.548791358 \\
 4 & 1.548591418 \\
 5 & 1.548591331
\end{array}
\right)$$ which are the solutions for ten significant figures.
For sure, these are the same as those given in Rhys Hughes's answer.
