Find the all explicit real roots of $x^{1/x}\ln x^2=x.$

Question

Using standard mathematical functions, find the all real roots of the equation: $$x^{1/x}\ln x^2=x.$$

I saw this question in the group of students studying mathematics. I tried to solve the equation with Lambert $W$ function but failed.

My attempt:

$$x^{1/x}\ln x^2=x\\ 2x^{1/x}\ln x=x$$

I tried the substitution $x=e^t$, then the equation become to

$$2te^{te^{-t}}=e^t$$

Now, how can I proceed?

Here is the Wolfram plot:

We have exactly $2$ real roots.

$$x ≈ 3.78195504319282$$

and

$$x ≈ 2.07412734548219$$

Answer 1

I do not think that we could have explicit solutions for the roots.

However, if you notice that the maximum of the function is at $x=e$, for an approximation, perform a series expansion.

You should have $$x^{\frac{1}{x}-1} \log \left(x^2\right)-1=\left(2 e^{\frac{1}{e}-1}-1\right)-e^{\frac{1}{e}-4} (1+e) (x-e)^2+O\left((x-e)^3\right)$$ Solving the quadratic gives $$x=e \pm \sqrt{ \frac{2 e^3-e^{4-\frac{1}{e}}}{1+e}}$$ that is to say $1.9185$ and $3.518$.

Now, polish the roots using Newton method.

Edit

In terms of approximation, we could have better building the $[2,n]$ Padé approximant of the lhs. So, to stay with quadratics, solve for $x$ $$1=\frac {a_0+a_1\,(x-e)+a_2\,(x-e)^2}{1+b_1\,(x-e)+b_2\,(x-e)^2}$$ where $$a_0=2 e^{\frac{1}{e}-1}\qquad a_1=\frac{10}{3} e^{\frac{1}{e}-2}\qquad a_2=\frac{e^{\frac{1}{e}-4} \left(-9+4 e+4 e^2\right)}{18 (1+e)}$$ $$b_1=\frac{5}{3 e}\qquad b_2=\frac{9+40 e+22 e^2}{36 e^3 (1+e)}$$ This gives $2.07430$ and $3.78243$.

Edit

After what I wrote in comments, as Snoopy uses to say :"Guess what ? I am happy" to have been wrong.

After @Bob Dobbs's nice answer, then the roots are $$x_1=\exp\Bigg[-W_0\left(-W_0\left(\frac{1}{2}\right)\right) \Bigg]=2.07413$$ $$x_2=\exp\Bigg[-W_{-1}\left(-W_0\left(\frac{1}{2}\right)\right) \Bigg]=3.78196$$

Answer 2

$x^{\frac{1}{x}}\ln x^2=x$ gives $\frac{\ln x}{x}e^{\frac{\ln x}{x}}=\frac{1}{2}$. Thus $\frac{\ln x}{x}=W(\frac{1}{2})$. Let $x=e^t$ then $-te^{-t}=-W(\frac{1}{2})$. So, $t=-W(-W(\frac{1}{2}))$ and $x=\exp(-W(-W(\frac{1}{2})))$ where $W$ is Lambert W Function.

You may find roots by using this calculator: https://www.had2know.org/academics/lambert-w-function-calculator.html