Given

$\rm 2^{4x + 5} \cdot a^{1 - x} = 4^{x + 3} \cdot 2 \cdot a ^{x -1}$

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = (2^{2})^{x + 3} \cdot 2^{1} \cdot a^{x - 1}$

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

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = 2^{2 (x + 3)} \cdot 2^{1} \cdot a^{x - 1}$

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = 2^{2x + 6} \cdot 2^{1} \cdot a^{x - 1}$

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

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = { 2^{(2x + 6) + 1} }\cdot a^{x - 1}$

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = 2^{2x + 7} \cdot a^{ x- 1}$

Due to the equality sign, the quantity on the left-hand side of the equation must be equal to the quantity on the right-hand side of the equation. Additionally, each of the two terms on both sides of the equation has the same base, so their exponents must be the same to hold equality. Hence, we equate the exponents of 2 on each side, and we get,

$\rm {4x + 5} = {2x + 7}$

$\rm or, 4x - 2x + 5 = 2x - 2x + 7$

$\rm or, 2x + 5 = 7$

$\rm or, 2x + 5 - 5 = 7 - 5$

$\rm or, 2x = 2$

$\rm \therefore x = 1$

Similarly, when we equate the exponents of a on each side, we get,

$\rm {1 - x} = {x - 1}$

$\rm or, 1 - x = x - 1$

$\rm or, 1 + (1 - x) = 1 + (x - 1)$

$\rm or, 2 -x = x$

$\rm or, 2 - x + x = x + x$

$\rm or, 2  = 2x$

$\rm \therefore x = 1$

Hence, the required value of x is 1.