Exponent laws extended - Year 9 Mathematics

What you will learn

apply the zero exponent rule to algebraic expressions,
interpret and simplify expressions with negative integer exponents,
extend the product, quotient, and power-of-a-power laws to integer exponents,
simplify expressions involving multiple bases and integer exponents,
connect negative exponents to scientific notation for small numbers.

Worked example 0 Real-world example: shrinking bacteria

A bacterial culture is being treated with an antibiotic. Each hour, the population is divided by $10$ . If the initial count is $5 \times 10^8$ bacteria:

After $1$ hour: $5 \times 10^8 \times 10^{-1} = 5 \times 10^7$ .
After $2$ hours: $5 \times 10^8 \times 10^{-2} = 5 \times 10^6$ .
After $h$ hours: $5 \times 10^8 \times 10^{-h} = 5 \times 10^{8-h}$ .
After $8$ hours: $5 \times 10^{8-8} = 5 \times 10^0 = 5$ bacteria.

Key idea: negative exponents model division and decay just as naturally as positive exponents model multiplication and growth.

1. Recap: the Year 8 index laws

Index laws (from Year 8, positive integer exponents)

Product of powers

$a^m \times a^n = a^{m+n}$

Quotient of powers

$a^m \div a^n = a^{m-n}, \quad a \neq 0$

Power of a power

$(a^m)^n = a^{mn}$

Power of a product

$(ab)^n = a^n b^n$

Power of a quotient

$\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}, \quad b \neq 0$

These five laws were established for positive integers $m$ and $n$ . The goal of this topic is to show they still work when $m$ and $n$ are any integers (positive, zero, or negative).

2. Zero and negative exponents

The zero exponent

From the quotient rule: $\dfrac{a^n}{a^n} = a^{n-n} = a^0$ . But $\dfrac{a^n}{a^n} = 1$ , so:

Zero exponent

$a^0 = 1, \qquad a \neq 0.$

This applies to any expression in the base — not just single variables:

Worked example 1 Zero exponent with algebraic expressions

Simplify: $(3x^2y)^0$ and $5m^0$ .

$(3x^2y)^0 = 1$ — the entire expression raised to the power zero equals $1$ (provided $x, y \neq 0$ ).
$5m^0 = 5 \times 1 = 5$ — only $m$ is raised to the power zero; the $5$ is a coefficient.

Negative exponents

From the quotient rule: $\dfrac{a^3}{a^5} = a^{3-5} = a^{-2}$ . But also $\dfrac{a^3}{a^5} = \dfrac{1}{a^2}$ . So:

Negative exponent

$a^{-n} = \dfrac{1}{a^n}, \qquad a \neq 0.$

Equivalently, $\dfrac{1}{a^{-n}} = a^n$ — a negative exponent in the denominator moves the base to the numerator.

Worked example 2 Evaluating negative exponents

Evaluate: $2^{-3}$ and $\left(\dfrac{3}{4}\right)^{-2}$ .

$2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$ .
$\left(\dfrac{3}{4}\right)^{-2} = \left(\dfrac{4}{3}\right)^{2} = \dfrac{16}{9}$ .

For the second: a negative exponent on a fraction flips the fraction, then applies the positive exponent.

3. Simplifying algebraic expressions with integer exponents

All five index laws extend to integer exponents. Use them in the same way, but expect negative exponents in your answers — then convert to positive exponents at the end if required.

Worked example 3 Product rule with negative exponents

Simplify $x^4 \times x^{-7}$ .

Add the exponents:

$x^4 \times x^{-7} = x^{4+(-7)} = x^{-3} = \dfrac{1}{x^3}.$

Worked example 4 Quotient rule producing a negative exponent

Simplify $\dfrac{a^2 b^3}{a^5 b}$ .

Handle each base separately:

$\dfrac{a^2}{a^5} = a^{2-5} = a^{-3}, \qquad \dfrac{b^3}{b^1} = b^{2}.$

So $\dfrac{a^2 b^3}{a^5 b} = a^{-3} b^2 = \dfrac{b^2}{a^3}$ .

Worked example 5 Power of a power with negative exponents

Simplify $(m^{-2})^3$ .

Multiply the exponents:

$(m^{-2})^3 = m^{(-2) \times 3} = m^{-6} = \dfrac{1}{m^6}.$

Worked example 6 Combining multiple laws

Simplify $\dfrac{(2x^3)^{-2} \times 4x}{x^{-4}}$ .

$(2x^3)^{-2} = 2^{-2} \times x^{-6} = \dfrac{1}{4} x^{-6}$ .
Numerator: $\dfrac{1}{4} x^{-6} \times 4x = x^{-6} \times x^1 = x^{-5}$ .
Division: $\dfrac{x^{-5}}{x^{-4}} = x^{-5-(-4)} = x^{-1} = \dfrac{1}{x}$ .

4. Connection to scientific notation

Negative exponents are exactly what scientific notation uses for small numbers:

$0.00047 = 4.7 \times 10^{-4}.$

The exponent $-4$ tells you to divide $4.7$ by $10^4$ , which is the same as moving the decimal four places left.

Worked example 7 Scientific notation and exponent laws together

Simplify $\dfrac{6 \times 10^{-3}}{2 \times 10^{4}}$ and express in scientific notation.

Divide the coefficients: $\dfrac{6}{2} = 3$ .
Subtract the exponents: $10^{-3-4} = 10^{-7}$ .
Result: $3 \times 10^{-7}$ .

Practice

Fluency

Tier 1: basic skills

Evaluate $5^0$ .
Evaluate $(4x)^0$ where $x \neq 0$ .
Evaluate $3^{-2}$ .
Write $10^{-5}$ as a fraction.
Simplify $a^3 \times a^{-5}$ . Give the answer with a positive exponent.
Simplify $\dfrac{m^4}{m^7}$ . Give the answer with a positive exponent.
Simplify $(x^{-3})^2$ .
Simplify $(2a)^{-3}$ .
Evaluate $\left(\dfrac{1}{2}\right)^{-4}$ .
Write $0.000\,56$ in scientific notation.

Reasoning

Tier 2: mixed practice

Simplify $\dfrac{a^5 b^{-2}}{a^{-1} b^3}$ and write with positive exponents only.
Simplify $(3x^{-2}y)^3$ .
Simplify $\dfrac{(p^2 q^{-1})^3}{p^{-3} q^4}$ .
Show that $a^{-1} + b^{-1} \neq (a+b)^{-1}$ by substituting $a = 2$ and $b = 3$ .
Simplify $\dfrac{12 x^{-3} y^2}{4 x^2 y^{-1}}$ and express with positive exponents.
A virus measures $8 \times 10^{-8}$ m across. Express this in nanometres.
Simplify $\left(\dfrac{2a^3}{b^{-2}}\right)^{-1}$ .
If $x^{-2} = 9$ , find the value of $x^2$ .

Reasoning

Tier 3: explain and apply

Explain why $a^{-n} = \dfrac{1}{a^n}$ must be true if the quotient rule is to remain valid for all integer exponents.
Simplify $\dfrac{(5a^2 b^{-3})^2}{(a^{-1} b^2)^3}$ completely.
The intensity of light decreases with the square of the distance: $I = I_0 \times d^{-2}$ . If you triple the distance, by what factor does the intensity change?
A student writes $(x + y)^{-1} = x^{-1} + y^{-1}$ . Disprove this with a counterexample and explain the error.
Simplify $\dfrac{(2 \times 10^3)^{-2} \times (5 \times 10^{-1})}{10^{-4}}$ and give the answer in scientific notation.

Challenge

Reasoning

Harder reasoning

Simplify $\dfrac{a^{2n} \times a^{-n+1}}{(a^2)^{n-1}}$ where $n$ is a positive integer.
Prove that for any non-zero $a$ and integers $m, n$ : $(a^m)^{-n} = (a^{-m})^n = a^{-mn}$ .
A radioactive substance halves every year. Write the fraction remaining after $t$ years as a power of $2$ . After how many years is less than $\dfrac{1}{1000}$ of the substance left? (Hint: solve $2^{-t} < 10^{-3}$ .)
Simplify $\dfrac{(3x^{-2}y^3)^{-2} \times (9x^4 y^{-1})^2}{(xy)^{-4}}$ completely, writing the answer with positive exponents.

Answers

Answer key

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Tier 1

$1$ .
$1$ .
$\dfrac{1}{9}$ .
$\dfrac{1}{100\,000}$ .
$\dfrac{1}{a^2}$ . Method: $a^{3+(-5)} = a^{-2} = \dfrac{1}{a^2}$ .
$\dfrac{1}{m^3}$ . Method: $m^{4-7} = m^{-3}$ .
$x^{-6} = \dfrac{1}{x^6}$ . Method: $(-3) \times 2 = -6$ .
$\dfrac{1}{8a^3}$ . Method: $2^{-3} \times a^{-3} = \dfrac{1}{8} \times \dfrac{1}{a^3}$ .
$16$ . Method: $\left(\dfrac{1}{2}\right)^{-4} = 2^4 = 16$ .
$5.6 \times 10^{-4}$ .

Tier 2

$\dfrac{a^6}{b^5}$ . Method: $a^{5-(-1)} = a^6$ ; $b^{-2-3} = b^{-5}$ .
$\dfrac{27y^3}{x^6}$ . Method: $3^3 x^{-6} y^3 = 27 x^{-6} y^3$ .
$\dfrac{p^9}{q^7}$ . Method: numerator $p^6 q^{-3}$ ; division gives $p^{6-(-3)} q^{-3-4} = p^9 q^{-7}$ .
$a^{-1} + b^{-1} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$ . But $(a+b)^{-1} = \dfrac{1}{5} = 0.2$ . Since $\dfrac{5}{6} \neq \dfrac{1}{5}$ , the two expressions are not equal.
$\dfrac{3y^3}{x^5}$ . Method: $\dfrac{12}{4} = 3$ ; $x^{-3-2} = x^{-5}$ ; $y^{2-(-1)} = y^3$ .
$80$ nm. Method: $8 \times 10^{-8} \text{ m} = 80 \times 10^{-9} \text{ m} = 80$ nm.
$\dfrac{b^{-2}}{2a^3} = \dfrac{1}{2a^3 b^2}$ . Method: flip and apply positive exponent.
$x^2 = \dfrac{1}{9}$ . Since $x^{-2} = 9$ , take the reciprocal: $x^2 = \dfrac{1}{9}$ .

Tier 3

The quotient rule says $\dfrac{a^m}{a^n} = a^{m-n}$ . When $m < n$ , we get $a^{m-n}$ with a negative exponent. But direct cancellation gives $\dfrac{1}{a^{n-m}}$ . For these to be equal, $a^{-(n-m)}$ must equal $\dfrac{1}{a^{n-m}}$ , which means $a^{-k} = \dfrac{1}{a^k}$ .
$\dfrac{25a^4 b^{-6}}{a^{-3} b^6} = 25 a^{4-(-3)} b^{-6-6} = 25 a^7 b^{-12} = \dfrac{25a^7}{b^{12}}$ .
Intensity at distance $3d$ : $I_0 \times (3d)^{-2} = I_0 \times \dfrac{1}{9d^2} = \dfrac{1}{9} \times I_0 d^{-2}$ . The intensity decreases by a factor of $9$ .
Let $x = 1, y = 1$ . Then $(1+1)^{-1} = \dfrac{1}{2}$ , but $1^{-1} + 1^{-1} = 2$ . The error is distributing an exponent over addition — the power-of-a-product rule applies to multiplication, not addition.
$(2 \times 10^3)^{-2} = 2^{-2} \times 10^{-6} = \dfrac{1}{4} \times 10^{-6}$ . Multiply by $5 \times 10^{-1}$ : $\dfrac{5}{4} \times 10^{-7} = 1.25 \times 10^{-7}$ . Divide by $10^{-4}$ : $1.25 \times 10^{-7-(-4)} = 1.25 \times 10^{-3}$ .

Challenge

$\dfrac{a^{2n} \times a^{-n+1}}{a^{2n-2}} = \dfrac{a^{2n + (-n+1)}}{a^{2n-2}} = \dfrac{a^{n+1}}{a^{2n-2}} = a^{(n+1)-(2n-2)} = a^{3-n}$ .
$(a^m)^{-n}$ : multiply exponents gives $a^{m \times (-n)} = a^{-mn}$ . $(a^{-m})^n$ : multiply exponents gives $a^{(-m) \times n} = a^{-mn}$ . Both equal $a^{-mn}$ .
Fraction remaining after $t$ years: $\left(\dfrac{1}{2}\right)^t = 2^{-t}$ . Need $2^{-t} < 10^{-3}$ , i.e. $2^t > 1000$ . Since $2^{10} = 1024 > 1000$ , after $10$ years less than $\dfrac{1}{1000}$ remains.
$(3x^{-2}y^3)^{-2} = 3^{-2} x^4 y^{-6} = \dfrac{x^4}{9y^6}$ . $(9x^4 y^{-1})^2 = 81 x^8 y^{-2}$ . Product: $\dfrac{81 x^{12}}{9 y^8} = 9 x^{12} y^{-8}$ . Divide by $(xy)^{-4} = x^{-4} y^{-4}$ : $9 x^{12-(-4)} y^{-8-(-4)} = 9 x^{16} y^{-4} = \dfrac{9x^{16}}{y^4}$ .

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