= Solution
The logarithm must be expanded about two, not about one. Factor
$$
1+e^x=2\left(1+\frac x2+\frac{x^2}4+O(x^3)\right).
$$
Applying the supplied logarithm expansion to the bracket gives $\log2+(x/2+x^2/4)-\tfrac12(x/2)^2+O(x^3)$. Thus, through second order,
$$
\boxed{\log(1+e^x)=\log2+\frac x2+\frac{x^2}{8}+O(x^3).}
$$
In fact the cubic coefficient vanishes, since $\log(1+e^x)=x/2+\log[2\cosh(x/2)]$ and the second term is even; the <Maclaurin series> remainder can be sharpened to $O(x^4)$.
Back to article page
