 Solution

Write $u=1/T$. Then $\log \zeta =\mathrm{constant}-(3/4)\log u-T_{H}u$, giving local slope $d{\log }_{10}\zeta /du=-(T_H+3T/4)/\log 10$. At 5000–6000 K its magnitude is about $3.6\times 10^4$ K. The low-ionization tail in the original graph drops roughly one decade per $2.8\times 10^{-5}$ K${}^{-1}$, consistent with this value.

At $u=2.0\times 10^{-4}$ K${}^{-1}$, the graph gives $T\simeq 5000$ K and $\zeta \simeq 0.004$. Invert the full equation to get $n=C{T}^{3/2}{e}^{-R/(k_{B}T)}(1-\zeta )/{\zeta }^2$, yielding $\boxed{n\sim 10^{18}{\mathrm{m}}^{-3}}$. The neighboring tail markers give the same order; plotting precision and the squared dependence on the read ionization fraction preclude claiming many significant digits.