How to solve $x^\sqrt x=\frac{3x}{2}$?

Question

If $x^\sqrt x=\frac{3x}{2}$ then how to solve it? I have tried by taking $\log$ in both side but couldn't reach a solution. Do you have any logarithmic or any other solution?

Answer 1

Using Mathematica, by inspection there are two solutions. (There are no solutions for $x \geq 3$, because $x^\sqrt{x}$ grows much faster than $\frac{3}{2}x$. This is something you can prove, even though it might not be possible to solve for the roots.)

They look to be about 0.4 and 2.2, so using FindRoot with these guesses gives $\frac{9}{4}$ (as the above commenter said) and $0.361634$. By WolframAlpha, the latter solution does not have an obvious closed form.

Answer 2

Try this substitution:

Let $y = \sqrt{x}$.

Then $y^{2} = x$, and so, the original equation transforms to

$(y^{2})^y = \frac{3y^{2}}{2}$,

which, after some algebraic manipulation, yields

$y^{2y - 2} = \frac{3}{2}$;

Upon inspection, we see that $y = \frac{3}{2}$;

Whence---

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

Answer 3

Some of the correct ways to solve such an equation is to use a numerical iterative method such as Bisection Method or Newton's Method. The bisection method may be easier to use since the derivative required by the later method is rather unfriendly.