= Solution
Take the <gas pressure> $P$ to vanish at the exterior boundary and the atmosphere to be thin enough that <surface gravity of a star> is constant. With constant <electron-scattering opacity>, negligible radiative acceleration, and <hydrostatic equilibrium in optical depth>,
$$
\frac{dP}{d\tau}=\frac g\kappa,\qquad P=\frac g\kappa\tau,\qquad \boxed{\frac{d\log P}{d\tau}=\frac1\tau}.
$$
If the radiative force is retained, $dP/dr=-\rho(g-\kappa F/c)$ for <gas pressure>, and the same result holds with the constant positive effective gravity $g_{\rm eff}=g-\kappa F/c$. A nonzero exterior gas pressure instead gives $d\log P/d\tau=1/(\tau+\kappa P(0)/g_{\rm eff})$. Total gas-plus-radiation pressure has the nonzero radiative surface term from part (ii), so the advertised exact $1/\tau$ formula presumes gas pressure with zero exterior gas pressure.
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