Given,

$\rm 2^{x+3} + 2^{x+1} = 80$

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

$\rm or, 2^x \cdot 2^3 + 2^x \cdot 2^1 = 80$

$\rm or, 2^x \left ( 2^3 + 2^1 \right ) = 80$

$\rm or, 2^x \left ( 8 + 2 \right ) = 80$

$\rm or, 2^x \cdot 10 = 80$

Dividing both sides of the equation by 10, we get,

$\rm or, 2^x \cdot \frac{10}{10} = \frac{80}{10}$

$\rm or, 2^x = 8$

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

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

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

$\rm \therefore x = 3$

Hence, the required value of x = 3.