Given

$\rm 2^x - 2^{x-2} = 6$

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

$\rm or, 2^x - \frac{2^x}{2^2} = 6$

$\rm or, 2^x - \frac{2^x}{4} = 6$

Multiplying both sides of the equation by 4, we get,

$\rm or, 2^x \cdot 4 - \frac{2^x}{4} \cdot 4 = 6 \cdot 4$

$\rm or, 4 \cdot 2^x - 2^x = 24$

$\rm or,  2^x ( 4 - 1) = 24$

$\rm or, 2^x (3) = 24$

Dividing both sides of the equation by 3, we get,

$\rm or, 2^x \cdot \frac{3}{3} = \frac{24}{3}$

$\rm or, 2^x = 8$

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

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.