 Solution

Write $F(x)=\ln (e^{ax}-b)$. Direct differentiation, using the chain rule, givesThe stated parameter restrictions ensure that the logarithm is defined near $x=1$, and that $F^{\prime }(1)>0$ and $F^{\prime \prime }(1)<0$. With $h=x-1$, the Taylor series beginsIf $e^a-b\ne 1$, the constant term is nonzero, and the first two nonzero terms areThis covers both $a>\ln (1+b)$ and $0<a<\ln (1+b)$; the constant term is respectively positive or negative.

If $a=\ln (1+b)$, which is possible precisely with $0<a<\ln 2$ and $b=e^a-1$, the constant term vanishes. The first two nonzero terms are thenNo further case is needed, since the linear and quadratic coefficients cannot vanish under the given restrictions. All remainders use Big O notation.