where "10^x" means "ten to the x" and "10^-x" really should be written "10^(-x)".
Pretty straightforward to use but what is the magnitude of α Cen which is a binary system with VA=-0.003 and VB=+1.333 (ArXiv 0401606)?
We want VA+B - 0 = 2.5 log[L0/(LA+LB)]
where 0 magnitude corresponds to L0 which is in fact the magnitude zero point that nobody really seems to agree on... But that is no problem since we work with ratios and L0/(LA+LB) is just the reciprocal of
LA/L0=10^[0-0.4MA] plus LB/L0=10^[0-0.4MB] ie
and you can add in as many components as you wish.
Try combining two first magnitude stars: (note that 10^-0.4=1/2.512=0.398)
M(both)=-2.5log(1/2.512+1/2.512)=0.2474, about 0.25
You could have done that in your head! Two first magnitude stars are twice as bright as magnitude V=1, a factor of two is 0.75 magnitudes (less than 1, brighter is smaller) or V=0.25.
For αCenAB we have VAB=
which nudges it ahead of Arcturus (V=-0.04) in the bright star race!
Finally, if the magnitude difference is greater than 2.5 magnitudes you can use the small magnitude difference cheat you use to amaze your astronomer friends: a magnitude difference of 1% is close to 0.01 magnitudes. (Works since the magnitude base is close to e...ex~1+x....) Try a binary with A=5 and B=8. The V=8 is about 16x fainter than the V=5 so it would add about 6.25% to the brightness bringing the V=5 down to V=4.9375. Doing the calculation exactly results in VAB=4.9336.
Two brightness Here is Another way Stars: combined: ratios: to do two stars: M(A) M(B) M(A+B) A/B 1+B/A DM M(A)-DM 1 1 0.2474 1 2 0.752575 0.247425 3 3 2.2474 1 2 0.752575 2.247425 3 4 2.6361 2.511 1.398 0.363851 2.636149 7 9 6.8403 6.309 1.158 0.15973 6.84027 2 7 1.9892 100 1.01 0.010803 1.989197 5.7 8.2 5.596518 10 1.1 0.103482 5.596518 3.2 7.9 3.185781 75.86 1.013 0.014219 3.185781