Solution
= Solution
At half ionization, <electron> and neutral densities are each $n_0/2$, so the left side of Saha's equation is $n_0/2$. Its <logarithm> changes by $d\log n_0=(3/2+R/(k_BT'))d\log T'$. To first order,
$$
\boxed{\frac{\Delta n_0}{n_0}\simeq\left(\frac32+\frac R{k_BT'}\right)\frac{\Delta T'}{T'}.}
$$
This holds for small fractional changes and the supplied ground-state partition approximation, not an exact finite-difference identity.