Extending the Lindemann Weierstrass Theorem

What I want to do is tinker with the Lindemann Weierstrass Theorem so I can ask 'what is so special about the number $$e$$'? I'll state the theorem below.

Lindemann-Weierstrass Theorem (Baker's Reformulation) Let $$\alpha_1, \dots, \alpha_n \in \bar{\mathbb{Q}}$$ be distinct and let $$a_1, \dots, a_n \in \bar{\mathbb{Q}}$$ be non-zero. Where $$\bar{\mathbb{Q}}$$ refers to the set of algebraic numbers. Then $$\sum_{i=1}^n{a_i e^{\alpha_i}} \neq 0$$

Note that we really have a polynomial (of sorts) that we have evaluated at $$e$$.
So consider the set of polynomials evaluated at a variable $$x \in \mathbb{R}$$.

$$S_x=\bigg \{\sum_{i=1}^n{a_i x^{\alpha_i}}: \alpha_i \in \bar{\mathbb{Q}} \text{ distinct}, a_i\in \bar{\mathbb{Q}} \text{ non-zero}, n\in \mathbb{N} \bigg \}$$

Now we can rewrite the theorem above as saying: $$0\notin S_e$$ and actually with no effort at all we can extend the theorem for any algebraic number $$\beta$$:

$$0\notin S_{e^\beta}$$. But $$S_x$$ is countably many polynomials (They are only kinda polynomials: the exponents are algebraic) and each have finitely many zeros. So the set $$L=\{x \in \mathbb{R}: 0\in S_x\}$$ is countable as well. That is to say that most complex numbers enjoy the property of not satisfying any polynomials of this type.

And $$L$$ the solutions of some generalized polynomial is a nontrivial(shown below) countable superset of the algebraic numbers.

Questions Is there a name in the literature for $$L$$? Are there any expectations on what $$\mathbb{R} \backslash L$$ must be? Is it likely that say $$\ln(2)\in \mathbb{R} \backslash L$$? Does there exist any polynomial with algebraic coefficients and exponents which $$\ln(2)$$ satisfies? We should comment that it can be seen immediately that $$\bar{Q} \subseteq L$$ which is why we should focus on transcendental numbers.

• The case with $x=\ln 2$, $n=2$ boils down to something short and non-obvious: Is there an algebraic nonzero $\alpha$ such that $(\ln 2)^\alpha$ is algebraic? – user210229 Dec 9 '18 at 1:58
• I can answer only the part of the question (why e is special). It is a value at a rational point of a solution of a linear differential equation (y'=y) with rational coefficients. This is crucial for the proofs. – Alexandre Eremenko Dec 12 '18 at 14:48