Index laws with variables | Year 9

01 / LEARN

The idea, step by step

Coefficients and powers do different jobs

In $4 x^{3}$ , the coefficient 4 multiplies the variable power. When multiplying $4 x^{3}$ by $2 x^{2}$ , multiply the coefficients and separately combine the same-base powers: $8 x^{5}$ . Unlike bases, such as $x$ and $y$ , stay separate.

For a quotient, divide the coefficients and subtract exponents of matching variables. A variable appearing in the denominator must not be zero, even if it later cancels from the simplified expression.

Power every factor in brackets

In ${(3 x^{2} y)}^{2}$ , the outer square affects 3, $x^{2}$ and $y$ : $9 x^{4} y^{2}$ . It is not enough to square only the final letter.

Use negative powers when they help the calculation, then write an answer with positive exponents if requested. For example, $x^{- 3} y^{2} = \frac{y^{2}}{x^{3}}$ for $x \neq 0$ . The zero-exponent rule also needs a non-zero base.

A common trap

Combining unlike variable bases or forgetting to raise the coefficient.

Instead: Only matching bases share exponent laws, and an outside power acts on every factor inside the brackets.

02 / WORKED EXAMPLES

See the thinking

Example 1

A quotient of variable powers

Simplify $\frac{12 x^{5} y^{2}}{3 x^{2} y}$ for $x, y \neq 0$ .

Divide the coefficients: $12 \div 3 = 4$ .
Subtract matching exponents: $x^{5 - 2} = x^{3}$ and $y^{2 - 1} = y$ .
The result is $4 x^{3} y$ , with the original non-zero restrictions.

4 x^{3} y

Example 2

Power of a product with a negative exponent

Write ${(2 a^{- 2} b^{3})}^{2}$ with positive exponents, for $a \neq 0$ .

Square each factor: $2^{2} a^{- 4} b^{6} = 4 a^{- 4} b^{6}$ .
$a^{- 4} = \frac{1}{a^{4}}$ .
The result is $\frac{4 b^{6}}{a^{4}}$ .

\frac{4 b^{6}}{a^{4}}

Check your understanding

Simplify $3 x^{2} \times 4 x^{- 1}$ for $x \neq 0$ .

Show answer

$12 x$

The coefficients give 12 and the exponents give $2 + (- 1) = 1$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Simplify $x^{3} x^{5}$ .
1 mark
Simplify $2 a^{2} \times 5 a^{4}$ .
2 marks
Simplify ${(3 x^{2})}^{3}$ .
2 marks
Write $\frac{8 p^{2}}{4 p^{5}}$ with positive exponents, for $p \neq 0$ .
2 marks
Simplify $\frac{15 x^{4} y^{3}}{5 x y^{2}}$ for $x, y \neq 0$ .
2 marks
A student writes ${(2 x y)}^{3} = 2 x^{3} y^{3}$ . Correct the answer and explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Simplify $x^{3} x^{5}$ .
1 mark
$x^{3 + 5} = x^{8}$ .
$x^{8}$
Simplify $2 a^{2} \times 5 a^{4}$ .
2 marks
$2 \times 5 = 10$ and $a^{2 + 4} = a^{6}$ .
$10 a^{6}$
Simplify ${(3 x^{2})}^{3}$ .
2 marks
$3^{3} = 27$ and ${(x^{2})}^{3} = x^{6}$ .
$27 x^{6}$
Write $\frac{8 p^{2}}{4 p^{5}}$ with positive exponents, for $p \neq 0$ .
2 marks
The coefficient is 2 and $p^{2 - 5} = p^{- 3}$ .
$2 p^{- 3} = \frac{2}{p^{3}}$ .
$\frac{2}{p^{3}}$
Simplify $\frac{15 x^{4} y^{3}}{5 x y^{2}}$ for $x, y \neq 0$ .
2 marks
$15 \div 5 = 3$ , $x^{4 - 1} = x^{3}$ , $y^{3 - 2} = y$ .
$3 x^{3} y$
A student writes ${(2 x y)}^{3} = 2 x^{3} y^{3}$ . Correct the answer and explain.
3 marks
The 2 is inside the brackets, so it must also be cubed.
$2^{3} x^{3} y^{3} = 8 x^{3} y^{3}$ .
$8 x^{3} y^{3}$

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum

Year 9 numerical integer exponent laws and their extension to variables; scientific notation and practical measurement applications. Excludes fractional-index and full surd algebra taught in Year 10.

Official curriculum source ↗

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