The provided truth table consists of three variables $A$, $B$, and $C$, and the output $X$. Let's analyze the table step by step to obtain the Boolean expression and simplify it:

Truth Table:

Steps to solve:

1. Write the minterms: The minterms are the rows where the output $X = 1$: $(A'B'C) + (A'BC) + (AB'C') + (ABC') + (ABC)$ Here, the apostrophe (') represents the negation of a variable.

2. Simplify the Boolean Expression: Let's group and simplify these minterms:

   • $(A'B'C) + (A'BC)$ can be simplified to $A'C$.
   • $(AB'C') + (ABC')$ can be simplified to $AC'$.
   • The term $ABC$ remains unchanged.

   So, the simplified Boolean expression is: $A'C + AC' + ABC$

3. Further Simplification: This expression cannot be simplified further without losing any essential terms.

Logic Diagram:

• The first term $A'C$ can be represented as a NOT gate for $A$, ANDed with $C$.
• The second term $AC'$ is $A$ ANDed with $C'$ (NOT gate for $C$).
• The third term $ABC$ is a simple AND gate for $A$, $B$, and $C$.

Finally, these terms are ORed together to produce the output $X$.

Would you like me to draw the logic diagram for this? Also, do you have any specific questions about this Boolean simplification process?

Here are 5 related questions for further exploration:

1. How can Karnaugh maps be used to simplify Boolean expressions?
2. What are the advantages of simplifying logic circuits in Boolean algebra?
3. How do AND, OR, and NOT gates interact in a real circuit?
4. Can we use De Morgan’s laws to simplify any of the terms further?
5. How would the truth table change if we introduced an additional variable?

Tip: Always look for common terms in Boolean expressions to group and simplify them more easily using laws like distribution and combination.