Show that run time complexity of operations on 2-4 tree is $O(\lg n)$
The tree is height-balanced, therefore the height of the tree is $O(\log n)$. So any insertion has to travel at most $O(\log n)$ steps deep into the tree to find the proper insertion place. Then it also takes another $O(\log n)$ steps the rebalance the tree. The same idea can be applied to searching and deleting in the tree. 

The $O(\log n)$ height bound may be difficult to see in a 2-3-4 tree, so I will offer an alternative way to come to this realization. Any 2-3-4 tree is isometric to a proper Red-Black tree. That is to say, "they are equivalent data structures. In other words, for every 2–3–4 tree, there exists at least one red–black tree with data elements in the same order." A Red-Black tree is a balanced binary search tree whose height is more clearly bounded by $O(\log n)$ and so are all of its operations.