Question: Bound the number of comparisons for searching in a 2-3 tree in the worst case. Do not use big-O notation.
Answer: It is easy to see that the depth of a 2-3 tree is in its worst case when every node has only one key. Then we have a complete binary tree over n keys, and the depth is lg (n+1) - 1. (I use lg for base-2 logarithms.) We make 2 comparisons for any node with two keys. So the bound is 2 (lg (n+1) - 1). The high-order term here is 2 lg n.
Compare with an AVL tre, where the answer is 1.44 lg n. (It's actually lg_phi n, where phi is the golden ratio, but 1.44 is about 1/lg phi.)
You may object that the above argument assumes for the worst-case depth that each node has 1 key for the worst-case comparisons per node that all nodes have 2 keys. This can in essence happen, if every node has 1 key except for those along the search path. In this case the depth will be lg (n+2) - 2, so an actual bound is 2 lg (n+2) - 4. (You might try verifying this; I may be wrong.) But this is the same high-order term, so I wouldn't worry about it too much.