[quote=@Kho]
[@BBeast] Is this true?!!

[img]http://img1.wikia.nocookie.net/__cb1489099833/uncyclopedia/images/math/1/a/9/1a9aec77827190cec6d85bd874d01ecd.png[/img]

If it is, I can't believe I've gone my whole life without this critical moooothematical truth.
[/quote]

The left hand side evaluates to
[center]log[sub]moo[/sub](Cow[sub]A[/sub])+log[sub]moo[/sub](Cow[sub]B[/sub])=log[sub]moo[/sub](Cow[sub]A[/sub]*Cow[sub]B[/sub]).[/center]

I can change the base of the right hand side to be the same base as the left hand side. I get
[center]log[sub]2Cow[/sub](Moo)=log[sub]moo[/sub](2Cow)/log[sub]moo[/sub](Moo).[/center]
If we assume that moo=Moo, then it simplifies to
[center]log[sub]2Cow[/sub](Moo)=log[sub]moo[/sub](2Cow).[/center]

From inspection, we can see that your equation is only true if 2Cow=Cow[sub]A[/sub]*Cow[sub]B[/sub]. 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[sup]2[/sup]. This then comes with the requirement that Cow[sub]A[/sub]*Cow[sub]B[/sub]=Cow[sup]2[/sup]. This is trivially true for the case Cow[sub]A[/sub]=Cow[sub]B[/sub], but is true in other cases too. And units are preserved.