Does the second positive solution (besides $x = 1$) of the equation $e^{x^2-1}=x^3-x\ln x$ have a closed form?

Question

Does the equation $$e^{x^2-1}=x^3-x\ln x$$ have a closed form solution ?

The given equation has $2$ positive real roots. Graphically

It is not hard to see that $x=1$ is a rational solution. The difficulty lies in finding the second irrational root. Wolfram Alpha gives an approximate solution for the second root: $$x\approx 0.45076365201730712954...$$

The notion that the second root is irrational is my assumption. Presumably, this root is irrational.

Questions.

How do we know if an irrational root has a closed form? If this solution does not have closed form, with what theory can we prove it ?

Clarification: By closed form it is meant the elementary functions and the Lambert W function.

Answer 1

Taking logarithm on both side of $$ e^{x^2-1}=x^3-x \ln x, $$ we have $$ x^2-\ln x=1+\ln \left(x^2-\ln x\right)\iff y=1+\ln y \tag*{(*)} , $$ where $y=x^2-\ln x.$

Since $$y=1+\ln y \iff y=1,$$ therefore \begin{aligned} x^2-\ln x =1 \Rightarrow & e^{x^2}=x \cdot e \\ \Rightarrow & e^{-1}=x e^{-x^2} \\ \Rightarrow & e^{-2}=x^2 e^{-2 x^2} \\ \Rightarrow & -2 e^{-2}=-2 x^2 e^{-2 x^2} \\ \Rightarrow & -2 x^2=W\left(-2 e^{-2}\right) \\ \Rightarrow & x=\sqrt{-\frac{W\left(-2 e^{-2}\right)}{2}} \doteq 0.450764 \end{aligned}

where $W(.)$ is the Lambert function.

Answer 2

$$e^{x^2-1}=x^3-x\ln x$$ We can take the logarithm on both sides. $$x^2-1=\ln x+\ln(x^2-\ln x)\\ x^2-\ln x=1+\ln(x^2-\ln x)$$ We can make a substitution. \begin{eqnarray} u&=&x^2-\ln x\\ e^u&=&\frac{e^{x^2}}x\\ xe^{-x^2}&=&e^{-u}\\ x^2e^{-2x^2}&=&e^{-2u}\\ -2x^2e^{-2x^2}&=&-2e^{-2u}\\ -2x^2&=&W(-2e^{-2u})\\ x&=&\sqrt{\frac{-W(-2e^{-2u})}2} \end{eqnarray} Now, solve for $u.$ \begin{eqnarray} u&=&1+\ln u\\ \ln u-u&=&-1\\ ue^{-u}&=&e^{-1}\\ -ue^{-u}&=&-e^{-1}\\ u&=&-W(-e^{-1})\\ x&=&\sqrt{\frac{-W(-2e^{2W(-e^{-1})})}2} \end{eqnarray} There is an infinite nunber of branches of the Lambert function. Only two suffice for real inputs. Here is the non-trivial value, using the $0$ branch, $W_0(\cdot)$. And here is the trivial value using the $-1$ branch, $W_{-1}(\cdot).$