Probably, showing $ \log(a)-\log(r) \neq n$ by induction, if possible.

I mean, $ \log(a)-\log(r) \neq n- s_2(n)-2 $

Sure, sounds sufficient to me. But you haven't even defined what $a$ and $r$ are. Are they all possible decompositions of the odd part of $n!$ into two factors? It certainly seems unlikely that $\log a - \log r > n$ will uniformly hold over all choices of $a$ and $r$. It also is certainly unlikely that $\log a - \log r < n-s_2(n)-2$ will uniformly hold. So what makes you believe there is any inductive structure on which to establish the inequality?

@ErickWong , u r correct about $a, r$ and , yes, $\log a - \log r > n$ or $\log a - \log r < n-s_2(n)-2$ will not uniformly hold over all choices/all possible decompositions of $a$ and $r$, but it seems quite possible to show that either $\log a - \log r < n-s_2(n)-2$ or $\log a - \log r > n-s_2(n)-2$ , in other words, to show $ \log(a)-\log(r) \neq n- s_2(n)-2 $, $\forall n>n_0$, for sufficient large $n_0$ , I am considering the possibility.