= Solution
Write $u=1/T$. Then $\log\zeta=\mathrm{constant}-(3/4)\log u-T_Hu$, giving local slope $d\log_{10}\zeta/du=-(T_H+3T/4)/\log10$. At 5000–6000 K its magnitude is about $3.6\times10^4$ K. The low-ionization tail in the original graph drops roughly one decade per $2.8\times10^{-5}$ K$^{-1}$, consistent with this value.
At $u=2.0\times10^{-4}$ K$^{-1}$, the graph gives $T\simeq5000$ K and $\zeta\simeq0.004$. Invert the full equation to get $n=C T^{3/2}e^{-R/(k_BT)}(1-\zeta)/\zeta^2$, yielding $\boxed{n\sim10^{18}\ \mathrm{m^{-3}}}$. The neighboring tail markers give the same order; plotting precision and the squared dependence on the read <ionization fraction> preclude claiming many significant digits.
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