Play with the Base

Suppose the number $N$ has a $k$ -digit representation in base $b$ . Then $b^{k - 1} \le N < b^k$ , so $k - 1 \le \log_b N < k$ .

The last inequality implies that $k - 1 = \lfloor \log_b N \rfloor$ , i.e. $k = \lfloor \log_b N \rfloor + 1$ .

Thus $p(n) = \lfloor \log_6 8^n \rfloor + 1 = \lfloor n \log_6 8 \rfloor + 1$ and $q(n) = \lfloor \log_4 6^n \rfloor + 1 = \lfloor n \log_4 6 \rfloor + 1$ . Then

$\begin{aligned} \lim_{n\rightarrow \infty} \dfrac{p(n)q(n)}{n^2} &= \lim_{n\rightarrow \infty} \dfrac{(\lfloor n \log_6 8 \rfloor +1)(\lfloor n \log_4 6 \rfloor +1)}{n^2} \\ \\ &= \lim_{n\rightarrow \infty} \left( \dfrac{\lfloor n \log_6 8 \rfloor +1}{n} \right) \left( \dfrac{\lfloor n \log_4 6 \rfloor +1}{n} \right) \\ \\ &= \lim_{n\rightarrow \infty} \left( \dfrac{\lfloor n \log_6 8 \rfloor}{n} + \dfrac{1}{n} \right) \left( \dfrac{\lfloor n \log_4 6 \rfloor}{n} + \dfrac{1}{n} \right) \\ \\ &= (\log_6 8) (\log_4 6) \qquad \color{#D61F06} \small \text{[**See note below**]} \\ \\ &= \log_4 8 = \boxed{\dfrac{3}{2}} \end{aligned}$

Note: in the above we made use of the fact that $\displaystyle\lim_{n\rightarrow \infty} \dfrac{\lfloor nx \rfloor}{n} = x$ . To see that this is true, first note that $\lfloor x \rfloor = x - \langle x \rangle$ , where $0 \le \langle x \rangle < 1$ . Then

$\begin{aligned} \displaystyle\lim_{n\rightarrow \infty} \dfrac{\lfloor nx \rfloor}{n} &= \displaystyle\lim_{n\rightarrow \infty} \dfrac{nx - \langle nx \rangle}{n} \\ &= x - \displaystyle\lim_{n\rightarrow \infty} \dfrac{\langle nx \rangle}{n} \quad = x \qquad \color{#3D99F6} \small \text{[since } 0 \le \langle nx \rangle < 1] \end{aligned}$