Given

$\rm 2^{x} + 2^{x + 1} + 2^{x + 2} + 2^{x + 4} = 62$

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

$\rm or, 2^{x} + 2^{x} \cdot 2^{1} + 2^{x} \cdot 2^{2} + 2^{x} \cdot 2^{4} = 62$

$\rm or, 2^{x} \cdot ( 1 + 2 + 2^{2} + 2^{4} ) = 62$

$\rm or, 2^{x} \cdot ( 1 + 2 + 4 + 16 ) = 62$

$\rm or, 2^{x} \cdot 23 = 62$

Dividing both sides of the equation by 23, we get,

$\rm or, 2^{x} \cdot \frac{23}{23} = \frac{62}{23}$

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

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

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

$\rm \therefore x = 1$

Hence, the required value of x is 1.