= Solution
Put $w=e^x-1=x+x^2/2+x^3/6+O(x^4)$. The local <logarithmic series> $\ln(1-w)=-w-w^2/2-w^3/3+O(w^4)$ gives
$$
\boxed{\ln(2-e^x)=-x-x^2-x^3+O(x^4).}
$$
For the infinite <Taylor series>, it is necessary to distinguish the real domain of the logarithm, $x<\ln2$, from its convergence interval about zero. Write $a=\ln2$. The complex zeros of $2-e^z$ are $a+2\pi i k$ for integers $k$, so the nearest <logarithmic singularity> to zero is at $a$. The <radius of convergence> is therefore $a$, giving convergence for $|x|<a$ and divergence for $|x|>a$.
The two endpoints need separate checking. Factor off the nearest singularity:
$$
h(z)=\ln(1-z/a)+H(z),\qquad H(z)=\ln\left(\frac{2-e^z}{1-z/a}\right),\qquad H(0)=0.
$$
The quotient has a nonzero removable value $2a$ at $z=a$; its other zeros are at $a+2\pi i k$ with $k\ne0$. Consequently it has a <holomorphic logarithm> on a zero-free disk larger than $|z|\le a$, which is a <simply connected domain>, and the <Taylor series> of $H$ converges absolutely at both real endpoints. At $x=-a$, the remaining series $-\sum_{n\ge1}(-1)^n/n$ converges by the <alternating series test>. At $x=a$, it is $-\sum_{n\ge1}1/n$ and diverges. Thus the full real convergence set is
$$
\boxed{-\ln2\le x<\ln2.}
$$
Convergence is absolute in the open interval and conditional at the left endpoint, where the sum equals $\ln(3/2)$. This endpoint distinction is an instance of <isolating a logarithmic singularity to test a power-series endpoint>.
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