Well, we could choose factorise it as log(50000 dollars) = log(50000 dollar^0.5 * 1 dollar^0.5) = log(50000 dollar^0.5) + log(1 dollar^0.5). That does keep the units of the addition operands the same. Now we only have to figure out what the log of a root-dollar is...
the logarithm of a dollar
It’s really just the same question again—why can’t I write log(1 dollar) = 0 (or maybe 0 dollar^0.5), the same as I would write log(1) = 0.
Well, we could choose factorise it as log(50000 dollars) = log(50000 dollar^0.5 * 1 dollar^0.5) = log(50000 dollar^0.5) + log(1 dollar^0.5). That does keep the units of the addition operands the same. Now we only have to figure out what the log of a root-dollar is...
It’s really just the same question again—why can’t I write log(1 dollar) = 0 (or maybe 0 dollar^0.5), the same as I would write log(1) = 0.
$1 = 100¢. Now try logging both sides by stripping off the currency units first!