@BBeast Is this true?!!
If it is, I can't believe I've gone my whole life without this critical moooothematical truth.
The left hand side evaluates to
logmoo(CowA)+logmoo(CowB)=logmoo(CowA*CowB).
I can change the base of the right hand side to be the same base as the left hand side. I get
log2Cow(Moo)=logmoo(2Cow)/logmoo(Moo).
If we assume that moo=Moo, then it simplifies to
log2Cow(Moo)=logmoo(2Cow).
From inspection, we can see that your equation is only true if 2Cow=Cow
A*Cow
B. This is probably not the case, unless one of the cows is equal to 2. Unless cows are dimensionless, then this 'rule' fails to hold up to dimensional analysis.
I would recommend the amendment that, instead of 2Cow, you use Cow
2. This then comes with the requirement that Cow
A*Cow
B=Cow
2. This is trivially true for the case Cow
A=Cow
B, but is true in other cases too. And units are preserved.