An old puzzle, extended precision - and logs

[sa-penguin]

I found an old [1907] puzzle book, Canterbury Puzzles One of the puzzles involved a cask holding 100 cups of wine. Every night, for 30 nights, a young monk steals a cup - and tops up the cask with a cup of water. The puzzle is to determine exactly how much wine the monk stole over those 30 nights.

This requires you to solve: 100 (1 - (99/100)^30) - which ends up having 60 decimal places.

While a quick Google search will find a arbitrary precision calculator, I was intrigued by the original 1907 answer: while giving the right value, it claimed the use of logarithms was required. I can see that:

log (0.99)^30 = 30 * log (0.99)

I just can't see how to calculate logs (and anti-logs) to the required number of decimal places. At best, in my youth, I saw tables of logarithms to 5 decimal places.

Is there a "lost art" for using log tables, to get the extended precision required? Or is there some other secret?