In simple terms, E=E0-alog(bc)=-a(log(b)+log(c))=-alog(b)-alog(c).
Let a=0.059 16/2; b=1/([H+]2+[H+]K1+K1K2), c=[H+]2FH₂A/FD[H+]2=FH₂A/FD.
bc={[H+]2FH₂A/([H+]2+[H+]K1+K1K2)}/{FD[H+]2}.
Hence:E=E0-(0.059 16/2)log{[H+]2FH₂A/([H+]2+[H+]K1+K1K2)}/{FD[H+]2} is the same as:
E=E0-(0.059 16/2)log(1/([H+]2+[H+]K1+K1K2))-(0.059 16/2)log(FH₂A/FD).