The sizes of list A and B are $m$ and $n$ respectively. It's given that $m>n$, which implies that $\log(m) > \log(n)$. Therefore, $m\log(m)>n\log(n)$.
My question is about the analysis of $m\log(n)$ vs. $n\log(m)$ given $m>n$.
I am looking for general cases and a general solution for this problem. I am not even sure if there is a general solution for this problem.
An example algorithm problem:
Given two sorted lists A & B of size $m$ and $n$. Find $a ∈ A$ and $b ∈ B$ such that $a + b = K$.
Algorithm 1: Keep picking an element $a ∈ A$ and perform a binary search, looking for $K-a$ in List B until a solution is found. The complexity is $\mathcal{O}(m\log(n))$
Algorithm 2: Keep picking an element $b ∈ B$ and Perform a binary search, looking for $K-b$ in List A until asolution is found. Complexity $\mathcal{O}(n\log(m))$
So, which Algorithm is better, 1 or 2?
I hope that the solution is not to compute $m\log(n)$ and $n\log(m)$ numerically and decide on the basis of that. I'm looking for a general solution.
EDIT: Since we are dealing with asymptotic analysis (complexity analysis), Please consider that m & n are very very large.
HINT
Diving both by $\log(m) \cdot \log (n)$ and analyze the function $$f(x) = \frac{x}{\log x}$$ to show if it is increasing or decreasing using $f'(x)$... You are looking for asymptotics, so more interested in large values of $x$, but in general this analysis can show you a point where behavior changes, if one exists.