Given,

$\rm 2^{x-4} = 4^{x-6}$

$\rm or, 2^{x-4} = (2^2)^{x-6}$

Using the law of indices: $\rm (a^b)^c = a^{bc}$

$\rm or, 2^{x-4} = 2^{2(x-6)}$

The base of the terms in the left-hand and right-hand sides of the equations are the same; hence, we equate their exponents.

$\rm or, x - 4 = 2(x-6)$

$\rm or, x - 4 = 2x - 12$

$\rm or, x - x -4 = 2x - x - 12$

$\rm or, -4 = x - 12$

$\rm or, -4 + 12 = x - 12 + 12$

$\rm or, 8 = x$

$\rm \therefore x = 8$

Hence, the required value of x = 8.