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The problem asks to express the given numbers as powers of 2. Let's go through each part:

**a) \( 4^2 \):**
- \( 4 = 2^2 \), so \( 4^2 = (2^2)^2 = 2^{4} \).

**b) \( 128^3 \):**
- \( 128 = 2^7 \), so \( 128^3 = (2^7)^3 = 2^{21} \).

**c) \( 256^2 \):**
- \( 256 = 2^8 \), so \( 256^2 = (2^8)^2 = 2^{16} \).

**d) \( 8^2 \):**
- \( 8 = 2^3 \), so \( 8^2 = (2^3)^2 = 2^6 \).

**e) \( 4^3 \):**
- \( 4 = 2^2 \), so \( 4^3 = (2^2)^3 = 2^6 \).

**f) \( 512^4 \):**
- \( 512 = 2^9 \), so \( 512^4 = (2^9)^4 = 2^{36} \).

**g) \( 32^3 \):**
- \( 32 = 2^5 \), so \( 32^3 = (2^5)^3 = 2^{15} \).

**h) \( 64^2 \):**
- \( 64 = 2^6 \), so \( 64^2 = (2^6)^2 = 2^{12} \).

**i) \( 128^4 \):**
- \( 128 = 2^7 \), so \( 128^4 = (2^7)^4 = 2^{28} \).

**j) \( 256^4 \):**
- \( 256 = 2^8 \), so \( 256^4 = (2^8)^4 = 2^{32} \).

**k) \( 16^5 \):**
- \( 16 = 2^4 \), so \( 16^5 = (2^4)^5 = 2^{20} \).

**l) \( 1024^5 \):**
- \( 1024 = 2^{10} \), so \( 1024^5 = (2^{10})^5 = 2^{50} \).

Each expression has been rewritten with a base of 2.

Would you like more details on any of these steps?

Here are five related questions for further practice:
1. How do you express \( 64^3 \) as a power of 2?
2. Simplify \( 512^5 \) as a power of 2.
3. Convert \( 16^4 \) into a power of 2.
4. Express \( 1024^3 \) with a base of 2.
5. How can you rewrite \( 4^4 \) as a power of 2?

**Tip:** When expressing numbers as powers of 2, start by finding how the base can be written as \( 2^k \), and then use exponent rules to simplify.

Laipsnį pakeiskite laipsniu, kurio pagrindas 2:

Learn how to express numbers as powers of 2 with detailed solutions for problems such as 4^2, 128^3, and 256^2. This exercise focuses on the laws of exponents and is suitable for students in Grades 9-11.

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