Is the real solution of $\ln(x)=-e^x$ transcendental?

The function $$f(x)=\ln(x)+ e^x$$ defined on $\mathbb R_+$ has a unique real root $u$. It satisfies $$\ln(u)=-e^u$$ and $$\frac{1}{u}=e^{e^u}$$ The numerical value is $$0.269874137573449223877\cdots $$

Can $u$ be proven to be transcendental ?

• Numerical analysis with PARI/GP with the algdep-command reveals large coefficients upto degree $50$ indicating that $u$ is probably transcendental.

• The Lindemann-Weierstrass theorem does not help , $e^{e^u}$ need not be transdencental for algebraic $u$. Perhaps, someone can prove that $u$ is transcendental assuming Schanuel's conjecture.

• If a transcendentality proof is not possible undonditionally, can at least the irrationality be proven unconditionally ?

This constant is definitely transcendental, assuming Schanuel's conjecture. Indeed, let us consider two cases. If $u,e^u$ are linearly independent over $\mathbb Q$, we can consider the extension $\mathbb Q(u,e^u,e^u,e^{e^u})$ of $\mathbb Q$. Schanuel's conjecture says, then this extension has transcendence degree at least $2$. But if $u$ was algebraic, then so would be $e^{e^u}=\frac{1}{u}$, and then there would be only one transcendental element among $u,e^u,e^{e^u}$.

If $u,e^u$ are linearly dependent, we can instead use Lindemann-Weierstrass - if $u$ was algebraic, $e^u$ would be transcendental, so they couldn't be dependent.