Given

$\rm 2^{x + 1} - 2^{x} = 8$

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

$\rm or, 2^{x} \cdot 2^{1} - 2^{x} = 8$

$\rm or, 2^{x} \left ( 2^{1} - 1 \right ) = 8$

$\rm or, 2^{x} (2 - 1) = 8$

$\rm or, 2^{x} \cdot 1 = 8$

$\rm or, 2^{x} = 8$

$\rm or, 2^{x} = 2^3$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the required value of x is 3.