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?
Best Answer
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.