How can I get the expression of x?

Question

If there is $$x^2e^{A\sqrt x}=B$$ then what is the expression of $x$? If this cannot be solved, is there any approximation?

Answer 1

There isn't a really nice form except using lambert W.

$$x^2e^{A\sqrt x}=B$$

$$2\ln x+A \sqrt x=\ln B$$

The substitution $u=\sqrt{x}$ gives:

$$4\ln u+Au=\ln B$$

$$\ln u+\frac{A}{4}u=\frac{\ln B}{4}$$

$$ue^{\frac{A}{4}u}=e^{\frac{\ln B}{4}}$$

$$\frac{A}{4}ue^{\frac{A}{4}u}=\frac{A}{4}e^{\frac{\ln B}{4}}$$

$$\frac{A}{4}u=W(\frac{A}{4}e^{\frac{\ln B}{4}})$$

So assuming $A \neq 0$:

$$u=\frac{4}{A}W(\frac{A}{4}e^{\frac{\ln B}{4}})$$

$$x=(\frac{4}{A}W(\frac{A}{4}e^{\frac{\ln B}{4}}))^2$$

Newtons method should give you approximate solutions.

Going back to the expression:

$$2\ln x+A \sqrt x=\ln B$$

$$2\ln x+A \sqrt x-\ln B=0$$

So let $x_1>0$ be your choice and let:

$$x_{n+1}=x_n-\frac{2\ln x_n+ A \sqrt{x_n}-\ln B}{\frac{2}{x_n}+A\frac{1}{2\sqrt{x_n}}}$$

Then, if the sequence converges:

$$\lim_{n \to \infty} x_n=x$$

So $x_n$ for sufficiently large $n$ should give you a good approximation. If you pick $x_0$ that causes the sequence not to converge, pick another one.

Answer 2

From Travis's comment and Ahmed's answer, there are not in general a closed form so just approximations. A way is to use the graphics of $e^{a\sqrt x}$ and $\frac{b}{x^2}$ separately to see the intersection point. In the figure below you have an example corresponding to the values $(a,b)=(0.8,12.4)$ for which the (approximate) solution is $(2, 3.1)$