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How can one solve this equation for $x$?

$$1 = x (\ln x + \ln^2 x) $$

Can it be solved analytically?

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    My spider senses thinks not - though I could be wrong. – 2017-02-21
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    Are you interested in numerical solutions, if it cannot be solved analytically? For positive solutions $x\gt 0$ you might consider the substituion $x = e^t$. – 2017-02-21
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    I cannot solve it analytically but i appoximated using graph intesection and all that and the answer pops out to be nearly 1.556. ;) – 2017-02-21
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    with e.g. the Newton-Raphson method we find $$1.55913351652565445336$$ – 2017-02-21
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    In fact, this is most likely an open problem - it's almost certain that such an $x$ is transcendental, but a proof is likely to be impossible with current technology. – 2017-02-22
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    @StevenStadnicki By Baker's theorem, this number $x$ is transcendental. – 2017-06-07
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    @i707107 How do you get Baker's theorem to work here? It seems as though the presence of $\ln^2 x$ defeats that theorem (which is all about linear independence). This even gets called out in the Wikipedia article: "Baker's theorem grants us the linear independence over the algebraic numbers of logarithms of algebraic numbers. This is weaker than proving their algebraic independence. _So far no progress has been made on this problem at all_." (emphasis mine) – 2017-06-07
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    @StevenStadnicki We can go with proof by contradiction. Suppose $x$ is algebraic. The equation gives $$\ln x = \frac{-1\pm\sqrt{1+\frac4x}}2$$. This gives a nontrivial linear combination of logarithms of algebraic number and $1$ (e.g. $\{ 1, \ln x \}$). Then Baker's theorem implies that $\{ \ln x \}$ is linearly dependent set over $\mathbb{Q}$. This is a contradiction. – 2017-06-07
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    I have used the same idea on this problem: https://math.stackexchange.com/questions/1992922/three-other-transcendental-numbers/2312822#2312822 – 2017-06-07
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    @i707107 Ahhh, I see. Yeah, as far as I can tell that should work; thank you! – 2017-06-07

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My initial reaction was to think it could be solved using the Lambert W function. Unfortunately, it seems like one instead needs to use a generalization, the quadratic Lambert W function.

I'm using the notation from On the generalization of the Lambert W function by Mezo and Baricz. See also this question from MathOverflow, which is related.

Anyway, we want to solve: $$ 1=y\left[ \ln(y) + \ln(y)^2\right] $$ First, transform (as suggested in the comments) via $y=\exp(x)$: $$ 1 = e^x(x+x^2) $$ Now, the generalized Lambert W inverts $$ f(x)=e^x(x-p)(x-q) $$ And so it can be shown that the solution to $$ c^2a=e^x(x-cp)(x-cq) $$ is given by $$ W_G\left( \begin{matrix} cp & cq \\ \, & \, \end{matrix} \,; \; c^2a \right) $$ where $W_G$ is the generalized Lambert W.

So for your case $1=e^x(x-0)(x-(-1))$, we can say: $$ x = W_G\left( \begin{matrix} 0 & -1\\ \, & \, \end{matrix} \,; \; 1 \right) $$ which seems like an exceptionally simple case (I'm certain it can be better simplified in fact) of the function. So it's an analytic solution in some sense :D

The linked paper, in addition to others, looks at many properties of the function, so maybe those would be more helpful for analysis (and less likely than me to have made mistakes).

As an aside, one neat fact is that the Maclaurin series of $W_G$ is given by $$ W_G\left( \begin{matrix} p & q \\ \, & \, \end{matrix} \,; \; a \right) = p - \sum\limits_{n=1}^\infty \frac{1}{n!\,n} \left[ \frac{an\exp(-p)}{q-p} \right]^n \;\mathcal{B}_{n-1} \left( \frac{-2}{n(q-p)} \right) $$ where $\mathcal{B}_{k}$ are the Bessel polynomials: $$ \mathcal{B}_{n}(z)=\sum\limits_{k=0}^\infty \frac{(n+k)!}{k!(n-k)!} \left( \frac{z}{2} \right)^k $$ Thus (ignoring questions of convergence), we can write: $$ x= \sum\limits_{n=1}^\infty \frac{-1}{n!\,n} \left[ \frac{n}{-1} \right]^n \, \sum\limits_{k=0}^\infty \frac{(n-1+k)!}{k!(n-1-k)!} \left( \frac{1}{n} \right)^k $$ Unfortunately, a quick check in Mathematica suggests it is outside the convergence radius.