Given

$\rm 5 \cdot 4 ^{x+1} - 16^x = 64$

$\rm or, 5 \cdot 4^{x + 1} - (4^2)^x = 64$

By using the law of indices, $\rm (a^m)^n = (a^n)^m = a^{mn}$, we get,

$\rm or, 5 \cdot 4^{x+1} - (4^x)^2 = 64$

By using the law of indices, $\rm a^{m+n} = a^m \cdot a^n$, we get,

$\rm or, 5 \cdot 4^x \cdot 4^1 - (4^x)^2x = 64$

$\rm or, 20 \cdot 4^x - (4^x)^2 = 4^3$

Let us consider $\rm a = 4^x$, we get,

$\rm or, 20 \cdot a - (a)^2 = 64$

$\rm or, a^2 - 20 a + 64 = 0$

The above equation is quadratic in a. By using the quadratic formula, we get,

$\rm or, a^2 - (16 + 4) a + 64 = 0$

$\rm or, a^2 - 16a - 4a + 64 = 0$

$\rm or, a (a - 16) - 4 (a - 16) = 0$

$\rm or, (a - 4) (a - 16) = 0$

Either

$\rm or, (a - 4) = 0$

$\rm or, a = 4$

By our assumption, $\rm a = 4^x$. We substitute the value back in the equation, we get,

$\rm or, 4^x = 4$

$\rm or, 4^x = 4^1$

The base of the terms on both sides of the equation are the same, so we equate their powers; we get,

$\rm \therefore x = 1$

Or

$\rm or, (a - 16) = 0$

$\rm or, a = 16$

By our assumption, $\rm a = 4^x$. We substitute the value back in the equation, we get,

$\rm or, 4^x = 16$

$\rm or, 4^x = 4^2$

The base of the terms on both sides of the equation are the same, so we equate their powers; we get,

$\rm \therefore x = 2$

Hence, the required values of x = {1, 2}.