Comparing the efficiecy of 2 run times I have a heap with $n$ elements. 
$k$ represent a number that is the height of one of the elements in the tree.

I need to compare two run times and prove what i claim. 

The 2 run times are: 
$$
(1)O(\lg n \cdot (\lg n - k))
$$

And: 
$$
(2)O(2^k \cdot \lg n)
$$

Where: 
$$
\lg n - k
$$
Is a constant value.



For the first run time formula, its seems that we get: $O(\lg n)$ 

But how do i find the run time (and prove) for the second run time? 

Thanks.
$\log(n)-k$ is const $\implies$ $2^{\log(n)-k}$ is const $\implies O(2^{k} \cdot \log(n)) = O(2^{\log(n)-k} \cdot 2^{k} \cdot \log(n)) = O(2^{\log(n)} \cdot \log(n)) = O(n \cdot \log(n))$