An algebra problem by Naren Bhandari

$4^x \times 6^y = 48^{12}$

$\log\ (4^x \times 6^y) = 12\ \log 48$

$\log\ 2^{2x} + \log\ 6^y = 12\ \log (6 \times 2^3)$

$2x\ \log\ 2 + y\ \log\ 6 = 12\ (\log\ 6 + 3\ \log\ 2)$

$2x\ \log\ 2 + y\ \log\ 6 = 12\ \log\ 6 + 36\ \log\ 2$

$2x\ \log\ 2 = 36\ \log\ 2 \Longrightarrow 2x = 36 \Longrightarrow x = 18$

$y\ \log\ 6 = 12\ \log\ 6 \Longrightarrow y = 12$

$x + y = 30$

$\large 4^{x} \times 6^{y} = 48^{12} \Longrightarrow 2^{2x} \times (2 \times 3)^{y} = (2^{4} \times 3)^{12} \Longrightarrow 2^{2x + y} \times 3^{y} = 2^{48} \times 3^{12}$ .

Then by the Fundamental Theorem of Arithmetic we have that

$\large 2x + y = 48, y = 12 \Longrightarrow 2x = 36 \Longrightarrow x + y = 18 + 12 = \boxed{30}$ .

From $4^{\blue x}6^{\red y} = 48^{12} = (2^43)^{12} = 2^{48}3^{12} = 2^{36}6^{12} = 4^{\blue{18}}6^{\red{12}}$ , $\implies x = 18$ and $y = 12$ . Therefore, $x+y = \boxed{30}$ .