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Is $\ln(1+e^z)$ transcendent for all algebraic $z$?

I already know Schanuel's conjecture and Baker's theorem. But I don't know how they can help here.

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    In other words: Are there algebraic numbers $z,w$ with $e^z+1-e^w=0$? – 2017-02-26

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We have Lindemann's theorem:

For any distint algebraic numbers $\alpha_1,\ldots,\alpha_n$ and any non-zero algebraic numbers $\beta_1,\ldots,\beta_n$, we have $$\beta_1e^{\alpha_1}+\ldots+\beta_ne^{\alpha_n}\ne 0 $$

So assume $z$ and $w:=\ln(1+e^z)$ are algebraic. We have $e^z-e^w+e^0=0$, which the theorem allows only if $1,w,z$ are not distinct, i.e., in the following cases

  • $w=z$. But then $e^0=0$, which is absurd
  • $w=0$. But then $e^z=0$, which is absurd
  • $z=0$. But then $w=\ln 2$, which is transcendental (again by Lindemann from $2e^0-e^w=0$)