Index laws with variables | Year 9

01 / LEARN

Understand the idea.

Coefficients and powers do different jobs

A term such as $4 x^{3}$ has two parts: the numerical coefficient $4$ and the variable power $x^{3}$ . In a product, multiply the coefficients and apply exponent laws separately to each matching variable base. Unlike letters remain separate factors.

In a quotient, divide coefficients and subtract exponents of the same variable. Preserve restrictions from the original expression: a denominator containing $x$ requires $x \neq 0$ , even if every factor of $x$ later cancels.

Power every factor in brackets

An exponent outside a product affects every factor. For ${(3 x^{2} y)}^{2}$ , square $3$ , multiply the exponent of $x$ by $2$ , and square $y$ . The result is $9 x^{4} y^{2}$ , rather than $3 x^{2} y^{2}$ .

Negative exponents can appear during simplification. To finish with positive exponents, rewrite those powers as reciprocals: $x^{- 3} y^{2} = \frac{y^{2}}{x^{3}}$ for $x \neq 0$ . A zero power of a variable also requires that variable to be nonzero.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A quotient of variable powers

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

Worked solution

1
Separate the coefficient and matching bases.
$\frac{12 x^{5} y^{2}}{3 x^{2} y} = \frac{12}{3} \frac{x^{5}}{x^{2}} \frac{y^{2}}{y}$
The original restrictions x and y nonzero allow division by these factors.
2
Simplify each factor.
$\begin{aligned} \frac{12}{3} & = 4 \\ \frac{x^{5}}{x^{2}} & = x^{5 - 2} = x^{3} \\ \frac{y^{2}}{y} & = y^{2 - 1} = y \end{aligned}$
Subtract the denominator exponent from the numerator exponent for each variable.
3
Recombine and retain the restrictions.
$4 x^{3} y, x \neq 0, y \neq 0$
Simplification does not extend the allowed inputs of the original quotient.

Answer

$4 x^{3} y$

EXAMPLE 02

Power of a product with a negative exponent

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

Worked solution

1
Apply the outside square to every factor.
${(2 a^{- 2} b^{3})}^{2} = 2^{2} {(a^{- 2})}^{2} {(b^{3})}^{2}$
A power distributes across a product.
2
Evaluate the coefficient and multiply exponents.
$2^{2} {(a^{- 2})}^{2} {(b^{3})}^{2} = 4 a^{- 4} b^{6}$
A power of a power multiplies the inner and outer exponents.
3
Rewrite the negative power as a reciprocal.
$4 a^{- 4} b^{6} = \frac{4 b^{6}}{a^{4}}, a \neq 0$
Only the factor with the negative exponent moves into the denominator.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$12 x$

Multiply coefficients to get $3 \times 4 = 12$ , then add matching exponents: $2 + (- 1) = 1$ . Hence the result is $12 x$ , retaining $x \neq 0$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Coefficients and powers do different jobs

Power every factor in brackets

See the method in action.

A quotient of variable powers

Separate the coefficient and matching bases.

Simplify each factor.

Recombine and retain the restrictions.

Power of a product with a negative exponent

Apply the outside square to every factor.

Evaluate the coefficient and multiply exponents.

Rewrite the negative power as a reciprocal.

Try it yourself.

Check your understanding

Keep practising