I came across below expression where I needed to find the value of K where the maximum value will occur for the below function
$P=\binom{k}{2} \times 2^{n-k}$
I re-wrote it as
$P=2^{n-1} \times \{\frac{k^2-k}{2^k}\}$
differentiating it with respect to k
$\frac{dP}{dk}=2^{n-1} \times \Biggl( \frac{(2k-1)-(k^2-k)(log2^k)}{2^k} \Biggr)$
My query is, can I keep base of log in the above expression to 2?
I know I need to equate the above value to 0, to find the critical point.
Please help me further I am stuck here.