Grade 9 Exponent Rules Review Worksheets 2024 - Free Printable

Let's solve each problem step by step using the laws of exponents. We'll simplify each expression and write the final answer with positive exponents only.

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🔷 Laws of Exponents Used:

1. Power of a Power: $(a^m)^n = a^{m \cdot n}$
2. Product of Powers: $a^m \cdot a^n = a^{m+n}$
3. Quotient of Powers: $\frac{a^m}{a^n} = a^{m-n}$
4. Power of a Product: $(ab)^n = a^n b^n$
5. Power of a Quotient: $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$
6. Negative Exponent: $a^{-n} = \frac{1}{a^n}$, so move to denominator to make positive.
7. Zero Exponent: $a^0 = 1$

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✔ Solutions:

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#### ① $(a^4)^2$

Use Power of a Power:
$(a^4)^2 = a^{4 \cdot 2} = a^8$

✔ Answer: $a^8$

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#### ② $p^6 \cdot p^{14}$

Use Product of Powers:
$p^6 \cdot p^{14} = p^{6+14} = p^{20}$

✔ Answer: $p^{20}$

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#### ③ $\frac{p^7}{p^5}$

Use Quotient of Powers:
$\frac{p^7}{p^5} = p^{7-5} = p^2$

✔ Answer: $p^2$

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#### ④ $(z^2)^3$

Use Power of a Power:
$(z^2)^3 = z^{2 \cdot 3} = z^6$

✔ Answer: $z^6$

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#### ⑤ $\frac{q^{10}}{q^6}$

Use Quotient of Powers:
$\frac{q^{10}}{q^6} = q^{10-6} = q^4$

✔ Answer: $q^4$

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#### ⑥ $\frac{l^2}{l}$

This is $\frac{l^2}{l^1} = l^{2-1} = l^1 = l$

✔ Answer: $l$

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#### ⑦ $(x^3b)^4 (xb^6)^2$

First, apply Power of a Product to each term:

- $(x^3b)^4 = (x^3)^4 \cdot b^4 = x^{12} b^4$
- $(xb^6)^2 = x^2 \cdot (b^6)^2 = x^2 b^{12}$

Now multiply:
$$
x^{12} b^4 \cdot x^2 b^{12} = x^{12+2} b^{4+12} = x^{14} b^{16}
$$

✔ Answer: $x^{14} b^{16}$

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#### ⑧ $\left(\frac{a^2 b}{b^{-3} c^4}\right)^3 (a^{-3} b)^{-2}$

We'll simplify in parts.

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Part A: $\left(\frac{a^2 b}{b^{-3} c^4}\right)^3$

Simplify inside first:

Numerator: $a^2 b$
Denominator: $b^{-3} c^4$

So:
$$
\frac{a^2 b}{b^{-3} c^4} = a^2 \cdot \frac{b}{b^{-3}} \cdot \frac{1}{c^4} = a^2 \cdot b^{1 - (-3)} \cdot c^{-4} = a^2 b^4 c^{-4}
$$

Now raise to the 3rd power:
$$
(a^2 b^4 c^{-4})^3 = a^{6} b^{12} c^{-12}
$$

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Part B: $(a^{-3} b)^{-2}$

Apply exponent:
$$
= (a^{-3})^{-2} \cdot b^{-2} = a^{6} b^{-2}
$$

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Now multiply Part A and Part B:
$$
a^6 b^{12} c^{-12} \cdot a^6 b^{-2} = a^{6+6} b^{12-2} c^{-12} = a^{12} b^{10} c^{-12}
$$

Now write with positive exponents:
$$
= \frac{a^{12} b^{10}}{c^{12}}
$$

✔ Answer: $\frac{a^{12} b^{10}}{c^{12}}$

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#### ⑨ $\left(\frac{x^2 y^7}{x^{-2} y^4}\right)^2$

Simplify inside the parentheses first:

$$
\frac{x^2 y^7}{x^{-2} y^4} = x^{2 - (-2)} y^{7 - 4} = x^{4} y^{3}
$$

Now square it:
$$
(x^4 y^3)^2 = x^{8} y^{6}
$$

✔ Answer: $x^8 y^6$

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#### ⑩ $\left(\frac{p^3 q^5}{r^7}\right) \cdot \left(\frac{p^2 r^0 q^3}{p^4 r^2}\right)^3$

Break this into two parts.

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Part A: $\frac{p^3 q^5}{r^7}$ → leave as is for now.

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Part B: $\left(\frac{p^2 r^0 q^3}{p^4 r^2}\right)^3$

Note: $r^0 = 1$, so numerator becomes $p^2 \cdot 1 \cdot q^3 = p^2 q^3$

So:
$$
\frac{p^2 q^3}{p^4 r^2} = p^{2-4} q^3 r^{-2} = p^{-2} q^3 r^{-2}
$$

Now cube it:
$$
(p^{-2} q^3 r^{-2})^3 = p^{-6} q^9 r^{-6}
$$

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Now multiply Part A and Part B:

$$
\frac{p^3 q^5}{r^7} \cdot p^{-6} q^9 r^{-6} = p^{3 + (-6)} q^{5 + 9} r^{-7 + (-6)} = p^{-3} q^{14} r^{-13}
$$

Now convert to positive exponents:

$$
= \frac{q^{14}}{p^3 r^{13}}
$$

✔ Answer: $\frac{q^{14}}{p^3 r^{13}}$

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| ① | $a^8$ |
| ② | $p^{20}$ |
| ③ | $p^2$ |
| ④ | $z^6$ |
| ⑤ | $q^4$ |
| ⑥ | $l$ |
| ⑦ | $x^{14} b^{16}$ |
| ⑧ | $\frac{a^{12} b^{10}}{c^{12}}$ |
| ⑨ | $x^8 y^6$ |
| ⑩ | $\frac{q^{14}}{p^3 r^{13}}$ |

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