Worst case running time to search for an element in a balanced BST with $n2^n$ elements
You state that searching an element in a BST takes time $\log n$, but this is wrong on two counts. First, this is (roughly) the worst-case number of comparisons. Second, the running time itself isn't exactly $n$ but is roughly $C\log n$ for some constant $C$ that depends on the exact machine model (assuming that comparisons take constant time). For this reason, it is better to use $\Theta$ notation (look it up!) and state that the worst-case running time of element lookup in a BST is $\Theta(\log n)$, where $n$ is the number of elements in the BST.

In our case, the number of elements is $n2^n$, and so the running time is $\Theta(\log(n2^n)) = \Theta(\log n + n) = \Theta(n)$. Asymptotically, $n + \log n = \Theta(n)$, and so there is no need to explicitly mention $\log n$.