Power Of A Product Rule, Worksheet, And Examples - Free Printable

Let's solve each problem using the Power of a Product Rule, which states:

> $$(a \cdot b)^n = a^n \cdot b^n$$

This rule applies when you have a product inside parentheses raised to a power. You distribute the exponent to each factor inside the parentheses.

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1) $(3 \cdot 10)^2$

Apply the rule:
$$
(3 \cdot 10)^2 = 3^2 \cdot 10^2 = 9 \cdot 100 = 900
$$

✔ Answer: $900$

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2) $(9 \cdot 6)^7$

$$
(9 \cdot 6)^7 = 9^7 \cdot 6^7
$$

We don’t need to compute the actual values unless asked, so we leave it as:
✔ Answer: $9^7 \cdot 6^7$

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3) $(4 \cdot 5)^{10}$

$$
(4 \cdot 5)^{10} = 4^{10} \cdot 5^{10}
$$

✔ Answer: $4^{10} \cdot 5^{10}$

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4) $(a \cdot b)^{13}$

$$
(a \cdot b)^{13} = a^{13} \cdot b^{13}
$$

✔ Answer: $a^{13} \cdot b^{13}$

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5) $(a \cdot b)^{20}$

$$
(a \cdot b)^{20} = a^{20} \cdot b^{20}
$$

✔ Answer: $a^{20} \cdot b^{20}$

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6) $(8^2 \cdot 3^7)^5$

This is a product of two expressions raised to a power. Apply the rule:
$$
(8^2 \cdot 3^7)^5 = (8^2)^5 \cdot (3^7)^5
$$

Now use the Power of a Power Rule: $(a^m)^n = a^{m \cdot n}$

So:
- $(8^2)^5 = 8^{2 \cdot 5} = 8^{10}$
- $(3^7)^5 = 3^{7 \cdot 5} = 3^{35}$

✔ Answer: $8^{10} \cdot 3^{35}$

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7) $(4^5 \cdot 2^4)^8$

Apply the rule:
$$
(4^5 \cdot 2^4)^8 = (4^5)^8 \cdot (2^4)^8
$$

Use power of a power:
- $(4^5)^8 = 4^{5 \cdot 8} = 4^{40}$
- $(2^4)^8 = 2^{4 \cdot 8} = 2^{32}$

✔ Answer: $4^{40} \cdot 2^{32}$

*(Note: You could also write $4^{40}$ as $(2^2)^{40} = 2^{80}$, but unless asked, we keep it as $4^{40}$ for simplicity.)*

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8) $(4 \cdot y^6)^3$

Apply the rule:
$$
(4 \cdot y^6)^3 = 4^3 \cdot (y^6)^3
$$

Now simplify:
- $4^3 = 64$
- $(y^6)^3 = y^{6 \cdot 3} = y^{18}$

✔ Answer: $64 \cdot y^{18}$ or $64y^{18}$

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9) $(x^5 \cdot y^6 \cdot z^7)^2$

Apply the rule to all three factors:
$$
(x^5 \cdot y^6 \cdot z^7)^2 = (x^5)^2 \cdot (y^6)^2 \cdot (z^7)^2
$$

Now apply power of a power:
- $(x^5)^2 = x^{10}$
- $(y^6)^2 = y^{12}$
- $(z^7)^2 = z^{14}$

✔ Answer: $x^{10} \cdot y^{12} \cdot z^{14}$

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10) $(4^2 \cdot y^4 \cdot z^6)^3$

Apply the rule:
$$
(4^2 \cdot y^4 \cdot z^6)^3 = (4^2)^3 \cdot (y^4)^3 \cdot (z^6)^3
$$

Simplify each:
- $(4^2)^3 = 4^{2 \cdot 3} = 4^6$
- $(y^4)^3 = y^{12}$
- $(z^6)^3 = z^{18}$

Now, $4^6 = (2^2)^6 = 2^{12}$, but again, unless required, we can leave as $4^6$.

✔ Answer: $4^6 \cdot y^{12} \cdot z^{18}$

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $900$ |
| 2 | $9^7 \cdot 6^7$ |
| 3 | $4^{10} \cdot 5^{10}$ |
| 4 | $a^{13} \cdot b^{13}$ |
| 5 | $a^{20} \cdot b^{20}$ |
| 6 | $8^{10} \cdot 3^{35}$ |
| 7 | $4^{40} \cdot 2^{32}$ |
| 8 | $64y^{18}$ |
| 9 | $x^{10} \cdot y^{12} \cdot z^{14}$ |
| 10 | $4^6 \cdot y^{12} \cdot z^{18}$ |

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🔍 Key Concept Recap:

The Power of a Product Rule allows us to distribute an exponent over multiplication:
$$
(ab)^n = a^n \cdot b^n
$$
This works for numbers, variables, and even expressions with exponents already present.

Let me know if you'd like these simplified further or written in expanded form!