If we suppose that we have positive integers $k$, $c$, and $v$, can we find the $n$ that minimizes:
$$k^n \frac{\log{2^v}}{\log{v}}v^{\log_2{(k \cdot v \cdot c/n)}}$$
If we suppose that we have positive integers $k$, $c$, and $v$, can we find the $n$ that minimizes:
$$k^n \frac{\log{2^v}}{\log{v}}v^{\log_2{(k \cdot v \cdot c/n)}}$$
Since $\log$ is an increasing function, it's equivalent to minimize the logarithm of your expression, which is $$ n\log k - \frac{\log v}{\log2}\log n + \big\{ \log\frac{\log 2^v}{\log v} + \frac{\log v}{\log2} \log kvc \big\}; $$ and the last expression, being independend of $n$, is irrelevant. The derivative of this function of $n$ is simply $\log k - \frac{\log v}{\log2} \frac1n$, whose unique root (assuming $v>1$) is $\frac{\log 2}{\log v}\log k$; and the second derivative test shows that the function of $n$ is minimized there.