Initially there are m balls in one bag, and n in the other, where m, n > 0. Two different operations are allowed:
(a) Remove an equal number of balls from each bag;
(b) Double the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
Operation (b) is now replaced with (c):
(c) Triple the number of balls in one bag.
Is it always possible to empty both bags after a finite sequence of operations?
(a),(b): Double the number of balls in the bag that contains the smaller number, until that number becomes at least as big as the number of balls in the other (if $m=n$, this is omittted). This is possible since $m,n > 0$.
We now have 2 bags with $x$ and $y$ balls, satisfying $x\ge y > x/2$. Take now $2y-x$ balls from each bag, which is a positive number and at most as big as the smaller number of balls in the bags, as $y \ge 2y-x \Leftrightarrow 0 \ge y-x \Leftrightarrow x \ge y$.
This leaves $2x-2y$ balls in the bag that just had $x$ balls and $x-y$ balls in the bag that just had $y$ balls. Double the number of balls in the latter, then you have $2x-2y$ balls in both bags, and take them all out to get 2 empty bags.
(a),(c): Note that the total number of balls in both bags changes by an even amount for both operations (a) and (c). That means it is impossible to get to a total of $0$ balls in both bags from the starting position if $m+n$ is odd.