= Solution
For a <photon gas> in thermal and <chemical equilibrium>, <photon> number is not a conserved thermodynamic charge. Number-changing processes can create and absorb <photons> without a corresponding conserved-number constraint, so the equilibrium <photon chemical potential> is zero.
At $T=5\times10^9\,\mathrm K$, $k_{\rm B}T\simeq0.431\,\mathrm{MeV}$. The other light species are <electrons>, <Positrons>, three <neutrino> species and their antiparticles. <Neutrinos> are <ultrarelativistic>; <electrons> and <Positrons> are only moderately relativistic at this particular <temperature> because $m_ec^2=0.511\,\mathrm{MeV}$. The later simple counting formulas use the intended <ultrarelativistic> approximation for them. Baryons, muons and heavier particles are not an abundant relativistic bath at this <temperature>.
Pair equilibrium implies $\mu_{e^-}+\mu_{e^+}=2\mu_\gamma=0$. Approximate charge symmetry, with the net <electron> excess tied to the much smaller baryon asymmetry, then makes both <electron> and <Positron> <chemical potentials> small compared with $k_{\rm B}T$. For <neutrinos> one additionally assumes negligible lepton asymmetry. Their near-zero <chemical potentials> are not implied merely by their statistics. After <neutrino decoupling>, a distribution initially having negligible <chemical potential> retains a redshifted <Fermi-Dirac distribution>, with its own <temperature>.
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