Fern-Ching Lin ([Lin 1983]) and Timothy Chow ([Chow 1999]) asked, when the solutions of a transcendental equation of elementary functions can be elementary numbers.
My question is:
To which more general kinds of transcendental equations can Lin's theorem be extended or generalized?
$\mathbb{L}$ denotes the Liouvillian numbers (= Elementary numbers). The Elementary numbers are subdivded into the Explicit elementary numbers $\mathbb{E}$ and the Implicit elementary numbers.
Lin's theorem:
If Schanuel's conjecture is true and $P(X,Y)\in\overline{\mathbb{Q}}[X,Y]$ is an irreducible polynomial involving both $X$ and $Y$ and $P(z_0,e^{z_0})=0$ for some nonzero $z_0\in\mathbb{C}$, then $z_0$ is not in $\mathbb{L}$.
A corollary of Lin's theorem is the conclusion "then $z_0$ is not in $\mathbb{L}$ and not in $\mathbb{E}$", because $\mathbb{E}\subset\mathbb{L}$.
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[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
Theorem 1.
Let $E$ be an elementary function with $E\colon \mathrm{dom}(E)\subseteq\mathbb{C}\to\mathbb{C}$, $z\in \mathrm{dom}(E)$ and $E(z)\neq 0$.
If Schanuel's conjecture is true and $P(X,Y)\in\overline{\mathbb{Q}}[X,Y]$ is an irreducible polynomial involving both $X$ and $Y$ and $P(E(z),e^{E(z)})=0$ for some $z\in\mathbb{C}$, then $z$ is not in $\mathbb{L}$ and not in $\mathbb{E}$.
Proof.
Lin's theorem and the corollary mentioned in the question imply $E(z)$ is not in $\mathbb{L}$ and not in $\mathbb{E}$.
Because $E$ is an elementary function, $E(z)\in\mathbb{L}$ for all $z\colon z\in \mathrm{dom}(E) \land z\in\mathbb{L}$.
But because $E(z)$ is not in $\mathbb{L}$ and not in $\mathbb{E}$ as stated above, $z$ is not in $\mathbb{L}$ and not in $\mathbb{E}$. $\ \ \ \square$
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Let
$E,E_1$ be non-constant elementary functions,
$P\in\overline{\mathbb{Q}}[x,y]\setminus(\overline{\mathbb{Q}}[x]\cup\overline{\mathbb{Q}}[y])$ irreducible over $\mathbb{C}$,
$p\in\overline{\mathbb{Q}}[x]\setminus\overline{\mathbb{Q}}$,
$R\in\overline{\mathbb{Q}}(x,y)\setminus(\overline{\mathbb{Q}}[x]\cup\overline{\mathbb{Q}}[y])$ in reduced form, with numerator irreducible over $\mathbb{C}$,
$r\in\overline{\mathbb{Q}}(x)\setminus\overline{\mathbb{Q}}$.
$$P(E(z),e^{E(z)})=0,\ \ \ \ P(E(z),\ln(E(z)))=0$$ $$R(E(z),e^{E(z)})=0,\ \ \ \ R(E(z),\ln(E(z)))=0$$ $$E(R(E_1(z),e^{E_1(z)}))=E(0),\ \ \ \ E(R(E_1(z),\ln(E_1(z))))=E(0)$$