Note that you're allowed to put more than one ball in each pan.
Suppose you are given eight balls and you know that one ball is slightly heavier than all the others, which are of equal weight. You're also given a balancing scale with which you can compare the weight of balls by putting some in one pan and some in the other. What's the minimum number of weighings you need to establish which ball is the heavier one?
The answer is two. Split the balls in two groups of three and one group of two. Compare the two groups of three using the scale. If one is heavier than the other, pick two balls from the heavier group and compare them on the scale. If one of them is heavier than the other, then that's your answer. If they weigh in equally, then the heavier ball is the one you didn't pick.
If the two groups of three balls weigh in equally, then the heavier ball is among the group of two. You can compare these using one further weighing, which will give you your answer.
It's log base 3 of N ( number of balls )
We solve this problem on the basis that the balance scales give three possible results: left heavy, right heavy, equal. They don't actually work that way. The pivot is typically so sensitive that any slight imbalance will cause a left or right down situation. In order to solve the problem we have to suspend disbelief and pretend that the scales will actually balance. Maybe the bearing is a little sticky, or maybe we note that it doesn't fall very quickly - which we then interpret as a balanced state.