Year 9 - Exponent laws - Free practice

Year 9 Exponent Laws Worksheets & Practice Tests

Index laws extended to variables and longer expressions.

Quick answer

What are the exponent laws and how do you apply them to algebraic expressions?

The exponent laws let you simplify expressions with powers. The key rules are: product law $a^m \times a^n = a^{m+n}$ , quotient law $a^m \div a^n = a^{m-n}$ , power law $(a^m)^n = a^{mn}$ , and zero exponent $a^0 = 1$ . For example, $x^3 \times x^4 = x^7$ and $(2x^2)^3 = 8x^6$ .

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Worked examples

Year 9 exponent laws: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Product and quotient laws with variables
Simplify each expression.

(a) $x^3 \times x^5$

(b) $\dfrac{y^8}{y^3}$

(c) $3a^4 \times 2a^2$
Show worked solution
Key idea: When multiplying powers with the same base, add the exponents. When dividing, subtract the exponents.
1. (a) $x^3 \times x^5 = x^{3+5} = x^8$
2. (b) $\dfrac{y^8}{y^3} = y^{8-3} = y^5$
3. (c) Multiply the coefficients and apply the product law to the variable: $3a^4 \times 2a^2 = (3 \times 2)a^{4+2} = 6a^6$
Answer: (a) $x^8$ (b) $y^5$ (c) $6a^6$
Example 2 · Power-of-a-power and zero exponent
Simplify each expression.

(a) $(m^3)^4$

(b) $(2p^2)^3$

(c) $5x^0 + (3y)^0$
Show worked solution
Key idea: For a power of a power, multiply the exponents. Any non-zero base raised to the power of zero equals 1.
1. (a) $(m^3)^4 = m^{3 \times 4} = m^{12}$
2. (b) Apply the power to both the coefficient and the variable: $(2p^2)^3 = 2^3 \times p^{2 \times 3} = 8p^6$
3. (c) $x^0 = 1$ and $(3y)^0 = 1$ , so $5x^0 + (3y)^0 = 5(1) + 1 = 6$
Answer: (a) $m^{12}$ (b) $8p^6$ (c) $6$
Example 3 · Multi-step exponent problem in context
A rectangular garden bed has a length of $4b^4$ metres and a width of $3b$ metres.

(a) Write a simplified expression for the area of the garden bed in square metres.

(b) A second garden bed has area $(2b^2)^3$ square metres. Write a simplified expression for how many times larger the second garden bed's area is than the first garden bed's area.
Show worked solution
Key idea: Find each area using the exponent laws, then divide to find the ratio.
1. (a) Area of first garden bed: $4b^4 \times 3b = 12b^5$ m²
2. (b) Area of second garden bed: $(2b^2)^3 = 2^3 \times b^{2 \times 3} = 8b^6$ m²
3. Divide the second area by the first:
  $\frac{8b^6}{12b^5} = \frac{8}{12} \times b^{6-5} = \frac{2}{3}b$
4. The second garden bed's area is $\dfrac{2b}{3}$ times that of the first.
Answer: (a) $12b^5$ m² (b) $\dfrac{2b}{3}$ times larger

Common mistakes

Where students usually lose marks

Multiplying exponents instead of adding in the product law

$x^3 \times x^5 = x^{15}$

When multiplying powers with the same base, add the exponents, not multiply them. The rule is $x^m \times x^n = x^{m+n}$ , so $x^3 \times x^5 = x^8$ .

Forgetting to raise the coefficient to the power

$(2p^2)^3 = 2p^6$

The power outside the brackets applies to every factor inside, including the number. Raise the coefficient as well: $(2p^2)^3 = 2^3 \times p^6 = 8p^6$ .

Thinking x⁰ = 0 instead of 1

$5x^0 = 0$

Any non-zero base raised to the power zero equals 1, never 0. So $x^0 = 1$ and $5x^0 = 5 \times 1 = 5$ .

Subtracting coefficients instead of dividing when using the quotient law

$\dfrac{6a^6}{3a^2} = 3a^{6-2}$ but then writing $3a^3$ by subtracting coefficients too

Only the exponents are subtracted when using the quotient law. The coefficients must be divided: $\dfrac{6a^6}{3a^2} = \dfrac{6}{3} \times a^{6-2} = 2a^4$ .

Try one yourself

Your turn

Simplify $\dfrac{(m^3)^4}{m^5}$ , giving your answer as a power of $m$ .

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About this worksheet

Why we built it

Year 9 extends the exponent laws from Year 8 numbers to algebraic variables and longer expressions. These worksheets give your student practice mixing the laws in single questions - the format that catches them out on tests.

What's covered

Sub-skills your student will practise

✓Product, quotient and power-of-a-power laws applied to variables
✓Zero exponent in algebraic expressions
✓Negative exponents (preview)
✓Simplifying multi-step exponent problems
✓Mixed-rule questions with multiple bases

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Frequently asked

Questions parents ask about exponent laws

How are Year 9 exponent laws different from Year 8?

Year 9 extends the laws to algebraic variables and longer expressions. The same rules apply but students mix them in single questions: (2x^3)^2 x x^4 / x^5 needs three different laws.

What do students do when bases differ?

Index laws only work when bases match. 2^3 x 3^2 can't be simplified into a single power - it stays as 8 x 9 = 72. Students sometimes try to combine unlike bases and get it wrong.

What about negative exponents?

Year 9 introduces them as a preview: a^-n = 1/a^n. So 2^-3 = 1/8. They become more central in Year 10 once scientific notation arrives.

Is the zero exponent a Year 9 topic?

Yes - a^0 = 1 for any non-zero a. Introduced in Year 8, extended to variables in Year 9. Most students remember the rule but not why it's true (it follows from the division law).

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