Class 8 Exponents and Powers Worksheet - Free Printable

Let's solve each problem step by step.

---

Q.1 Express each of the following numbers as a product of powers of their prime factors:

#### a) 392
1. Start by dividing 392 by the smallest prime number, which is 2:
\[
392 \div 2 = 196
\]
2. Divide 196 by 2:
\[
196 \div 2 = 98
\]
3. Divide 98 by 2:
\[
98 \div 2 = 49
\]
4. Now, 49 is not divisible by 2, so try the next smallest prime number, which is 7:
\[
49 \div 7 = 7
\]
5. Finally, divide 7 by 7:
\[
7 \div 7 = 1
\]

So, the prime factorization of 392 is:
\[
392 = 2^3 \times 7^2
\]

#### b) 864
1. Start by dividing 864 by 2:
\[
864 \div 2 = 432
\]
2. Divide 432 by 2:
\[
432 \div 2 = 216
\]
3. Divide 216 by 2:
\[
216 \div 2 = 108
\]
4. Divide 108 by 2:
\[
108 \div 2 = 54
\]
5. Divide 54 by 2:
\[
54 \div 2 = 27
\]
6. Now, 27 is not divisible by 2, so try the next smallest prime number, which is 3:
\[
27 \div 3 = 9
\]
7. Divide 9 by 3:
\[
9 \div 3 = 3
\]
8. Finally, divide 3 by 3:
\[
3 \div 3 = 1
\]

So, the prime factorization of 864 is:
\[
864 = 2^5 \times 3^3
\]

Final Answer for Q.1:
\[
\boxed{392 = 2^3 \times 7^2, \quad 864 = 2^5 \times 3^3}
\]

---

Q.2 Choose the correct option:

#### a) \(3^4 \times 6^2 = 18^6\)
- Simplify \(6^2\):
\[
6^2 = (2 \times 3)^2 = 2^2 \times 3^2
\]
- So, \(3^4 \times 6^2\) becomes:
\[
3^4 \times (2^2 \times 3^2) = 2^2 \times 3^6
\]
- On the right-hand side, \(18^6\):
\[
18^6 = (2 \times 3^2)^6 = 2^6 \times 3^{12}
\]
- Clearly, \(2^2 \times 3^6 \neq 2^6 \times 3^{12}\). So, this option is false.

#### b) \(4^0 = 0\)
- Any non-zero number raised to the power of 0 is 1:
\[
4^0 = 1
\]
- So, this option is false.

#### c) \(5^2 > 25\)
- Calculate \(5^2\):
\[
5^2 = 25
\]
- Since \(25 = 25\), it is not greater than 25. So, this option is false.

#### d) \(((a^m)^n)^y = a^{mny}\)
- Using the property of exponents \((x^a)^b = x^{a \cdot b}\):
\[
((a^m)^n)^y = a^{m \cdot n \cdot y} = a^{mny}
\]
- This option is true.

Final Answer for Q.2:
\[
\boxed{d}
\]

---

Q.3 Simplify and write the answer in exponential form:

#### a) \((4^2)^3 \div 4^4\)
- Simplify \((4^2)^3\) using the property \((x^a)^b = x^{a \cdot b}\):
\[
(4^2)^3 = 4^{2 \cdot 3} = 4^6
\]
- Now, simplify \(4^6 \div 4^4\) using the property \(x^a \div x^b = x^{a - b}\):
\[
4^6 \div 4^4 = 4^{6 - 4} = 4^2
\]

#### b) \((5^5 \times 5^3) \times 5^1\)
- Use the property \(x^a \times x^b = x^{a + b}\):
\[
5^5 \times 5^3 = 5^{5 + 3} = 5^8
\]
- Now, multiply by \(5^1\):
\[
5^8 \times 5^1 = 5^{8 + 1} = 5^9
\]

#### c) \(10^0 \times 6^0\)
- Any number raised to the power of 0 is 1:
\[
10^0 = 1 \quad \text{and} \quad 6^0 = 1
\]
- So:
\[
10^0 \times 6^0 = 1 \times 1 = 1
\]
- In exponential form, \(1\) can be written as \(10^0\) or any other base raised to 0. For simplicity, we write:
\[
10^0
\]

Final Answer for Q.3:
\[
\boxed{4^2, 5^9, 10^0}
\]

---

Q.4 Express:

#### a) 729 as a power of 3
- Prime factorize 729:
\[
729 \div 3 = 243, \quad 243 \div 3 = 81, \quad 81 \div 3 = 27, \quad 27 \div 3 = 9, \quad 9 \div 3 = 3, \quad 3 \div 3 = 1
\]
- So, \(729 = 3^6\).

#### b) 128 as a power of 2
- Prime factorize 128:
\[
128 \div 2 = 64, \quad 64 \div 2 = 32, \quad 32 \div 2 = 16, \quad 16 \div 2 = 8, \quad 8 \div 2 = 4, \quad 4 \div 2 = 2, \quad 2 \div 2 = 1
\]
- So, \(128 = 2^7\).

#### c) 343 as a power of 7
- Prime factorize 343:
\[
343 \div 7 = 49, \quad 49 \div 7 = 7, \quad 7 \div 7 = 1
\]
- So, \(343 = 7^3\).

#### d) 256 as a power of 2
- Prime factorize 256:
\[
256 \div 2 = 128, \quad 128 \div 2 = 64, \quad 64 \div 2 = 32, \quad 32 \div 2 = 16, \quad 16 \div 2 = 8, \quad 8 \div 2 = 4, \quad 4 \div 2 = 2, \quad 2 \div 2 = 1
\]
- So, \(256 = 2^8\).

Final Answer for Q.4:
\[
\boxed{3^6, 2^7, 7^3, 2^8}
\]

---

Q.5 Which one is greater, 23 or 32?

- Compare the two numbers directly:
\[
23 < 32
\]

Final Answer for Q.5:
\[
\boxed{32}
\]

---

Final Answers:

\[
\boxed{392 = 2^3 \times 7^2, \quad 864 = 2^5 \times 3^3, \quad d, \quad 4^2, 5^9, 10^0, \quad 3^6, 2^7, 7^3, 2^8, \quad 32}
\]