Product of exponents

$6^x 14^y = 48 × 42^y$

$6^x = 48 × 3^y$

$2^x 3^x = 2^4 × 3 × 3^y$

$2^{x - 4} = 3^{y + 1 - x}$

Since both exponents are integers and 2 and 3 are coprimes, therefore the equation above can only stand, when both exponents are zero.

This gives us:

$x - 4 = 0 \Rightarrow x = 4$

and

$y + 1- x = 0 \Rightarrow y + 1 - 4 = 0 \Rightarrow y = 3$

Hence, our answer should be:

$\boxed { (4, 3) }$

Factoring the exponents, we get $2^x\cdot3^x\cdot2^y\cdot7^y = 2^4\cdot3^1\cdot2^y\cdot3^y\cdot7^y$ . Dividing by $7^y$ and combining equal-base exponents we get $2^{x+y}\cdot3^x = 2^{4+y}\cdot3^{1+y}$ . Looking at powers of 2, $x+y = 4+y$ , so $x = 4$ . Looking at powers of 3, $x = 1+y$ , so $y=3$ . Our ordered pair is $(4,3)$ .