$2.)$ A $3-$input XOR gate is equivalent to the circuit shown below:

The Boolean equation can be written as:

$x=(A^{\text{'}}B+A{B}^{\text{'}}{)}^{\text{'}}C+(A^{\text{'}}B+A{B}^{\text{'}})C^{\text{'}}$

Or it simply denoted as:

$x=Ao+Bo+C$

Using only AND, OR and NOT gates to implement the above Boolean equation, how many gates are

needed? Draw the logic diagram. Compare the savings of a single XOR gate implementation with the circuit

you just drew.

$($Please include drawings$)$