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Lesson 9 ~25 min Unit 1 · Index Laws +85 XP

Negative Indices — Evaluation

Evaluate fractional bases, combine negatives with positives, and write every final answer with positive indices only.

Today's hook: A negative index doesn't make a number negative — it makes it a FRACTION. And a fraction raised to a negative power flips into a whole number. How does that work?
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Printable Worksheets

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Think First
warm-up

Predict the value of $\left(\dfrac{1}{2}\right)^{-3}$. Will the answer be a fraction, a whole number, or something else? Then check by writing it as $\dfrac{1}{(1/2)^3}$.

Record in your workbook.
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The Big Idea
+5 XP

To evaluate a negative-index expression, reciprocate first, then evaluate using a positive index. Final answers always use positive indices.

$a^{-n} = \dfrac{1}{a^n}$ and $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$ — flip the fraction, then raise to the positive power. When combining with positive indices, add or subtract as normal, then convert any leftover negative to a positive index for the final answer.

$\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$
Flip fractions
$\left(\tfrac{1}{2}\right)^{-3} = 2^3 = 8$.
Combine indices
$a^{-3} \times a^5 = a^{-3+5} = a^2$.
Positive only
Always finish with positive indices in the answer.
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What You'll Master
objectives

Know

  • $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$
  • Combine negative + positive with $a^m \times a^n = a^{m+n}$
  • Final answers use positive indices

Understand

  • Why reciprocating a fraction "flips" it
  • Why a negative base with negative index keeps sign rules separate
  • Why "no negative indices" is standard form

Can Do

  • Evaluate $3^{-2}, 5^{-1}, \left(\tfrac{1}{2}\right)^{-3}$
  • Simplify $a^{-3} \times a^5$
  • Convert $\left(\tfrac{2}{3}\right)^{-2}$ to a positive answer
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Words You Need
vocabulary
ReciprocateTake $\dfrac{1}{\text{thing}}$; for fractions, flip top and bottom.
Standard formFinal answer with positive indices only.
Sign of base vs index$(-2)^{-3}$: the negative sign on the base is separate from the index.
Flip rule$\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$.
CombineApply product/quotient rule even when indices are negative.
Convert backUse $a^{-n} = \tfrac{1}{a^n}$ to remove negatives at the end.
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Spot the Trap
heads-up

Wrong: "$(-2)^{-3} = \dfrac{1}{8}$" — ignoring the negative base.

Right: $(-2)^{-3} = \dfrac{1}{(-2)^3} = \dfrac{1}{-8} = -\dfrac{1}{8}$. Negative base $\times$ odd index keeps the negative.

Wrong: "$\left(\tfrac{2}{3}\right)^{-2} = \tfrac{4}{9}$" — forgetting to flip the fraction.

Right: $\left(\tfrac{2}{3}\right)^{-2} = \left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4}$.

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Flipping Fractions
+5 XP

When the base is a fraction and the index is negative, flip the fraction and use the positive index.

$\left(\dfrac{a}{b}\right)^{-n} = \dfrac{1}{\left(\dfrac{a}{b}\right)^n} = \dfrac{1}{\dfrac{a^n}{b^n}} = \dfrac{b^n}{a^n} = \left(\dfrac{b}{a}\right)^n$. So flip the fraction, drop the minus.

$\left(\tfrac{1}{2}\right)^{-3} = 2^3 = 8$
6
Combining Indices
+5 XP

The product and quotient rules still work when indices are negative — just add or subtract carefully.

$a^{-3} \times a^5 = a^{-3+5} = a^2$. $a^{-2} \times a^{-3} = a^{-2-3} = a^{-5} = \dfrac{1}{a^5}$. If the sum stays negative, finish by writing as a fraction.

$a^{-3} \times a^5 = a^2$

Watch Me Solve It · 3 examples

Watch Me Solve It · Fraction with negative index
+15 XP per step
Q1
PROBLEM
Evaluate $\left(\dfrac{2}{3}\right)^{-2}$.
  1. 1
    Flip the fraction
    $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2$
    Drop the minus by reciprocating.
  2. 2
    Apply the index
    $\left(\dfrac{3}{2}\right)^2 = \dfrac{3^2}{2^2} = \dfrac{9}{4}$
  3. 3
    Final
    $\dfrac{9}{4}$
Answer$\dfrac{9}{4}$
Watch Me Solve It · Combine indices
+15 XP per step
Q2
PROBLEM
Simplify $a^{-3} \times a^5$ leaving your answer with a positive index.
  1. 1
    Apply product rule
    $a^{-3} \times a^5 = a^{-3+5}$
    Same base, add indices.
  2. 2
    Simplify the index
    $a^{2}$
  3. 3
    Check positive
    $a^2$ (already positive index)
    No conversion needed.
Answer$a^2$
Watch Me Solve It · Negative base, negative index
+15 XP per step
Q3
PROBLEM
Evaluate $(-2)^{-3}$.
  1. 1
    Apply negative-index rule
    $(-2)^{-3} = \dfrac{1}{(-2)^3}$
  2. 2
    Evaluate $(-2)^3$
    $(-2)^3 = -8$
    Negative $\times$ odd index = negative.
  3. 3
    Combine
    $\dfrac{1}{-8} = -\dfrac{1}{8}$
Answer$-\dfrac{1}{8}$

Common Pitfalls

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Common Pitfalls
heads-up
Forgetting to flip
$\left(\tfrac{2}{3}\right)^{-2} \ne \tfrac{4}{9}$. You must reciprocate first.
Fix: Flip the fraction, then apply the positive index.
Mixing up sign rules
A negative BASE and a negative INDEX do different things. $(-2)^{-3} = -\tfrac{1}{8}$, not $\tfrac{1}{8}$.
Fix: Treat the negative base and the negative index as separate steps.
Leaving a negative index
$a^{-5}$ is not a final answer in most NSW questions.
Fix: Convert to $\dfrac{1}{a^5}$ at the end — standard form uses positive indices.
Copy Into Your Books

Negative on a fraction

  • $\left(\tfrac{a}{b}\right)^{-n} = \left(\tfrac{b}{a}\right)^n$
  • Flip then raise
  • Drop the minus

Combine with positive

  • $a^{-3} \times a^5 = a^2$
  • Add indices as normal
  • Convert any negative at end

Negative base

  • $(-2)^{-3} = -\tfrac{1}{8}$
  • Base sign & index separate
  • Odd power keeps the minus

Final form

  • Positive indices only
  • $a^{-5} \to \dfrac{1}{a^5}$
  • Standard form

How are you completing this lesson?

Brain Trainer · 4 problems

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Brain Trainer · Evaluation drill
4 problems

Four drills mixing fractions, negative bases and combinations.

  1. 1 Evaluate $\left(\dfrac{1}{3}\right)^{-2}$.

    Flip to $3$, then square.$9$

  2. 2 Simplify $a^{-2} \times a^7$.

    $-2 + 7 = 5$.$a^5$

  3. 3 Evaluate $5^{-1}$ as a decimal.

    $\dfrac{1}{5}$.$0.2$

  4. 4 Simplify $b^{-4} \times b^{-2}$ with positive index.

    $-4 + (-2) = -6$, so $b^{-6}$.$\dfrac{1}{b^6}$
Complete in your workbook.

Quick Check · 5 questions

1
Evaluate $3^{-2}$.
+10 XP
2
Evaluate $\left(\dfrac{1}{2}\right)^{-3}$.
+10 XP
3
Simplify $a^{-3} \times a^5$.
+10 XP
4
Evaluate $\left(\dfrac{2}{3}\right)^{-2}$.
+10 XP
5
Evaluate $(-2)^{-3}$.
+10 XP

Show Your Working · 3 questions

Show Your Working
9 marks total
ApplyMedium3 MARKS

Q6. Evaluate each: (a) $5^{-1}$ as a decimal, (b) $\left(\dfrac{1}{4}\right)^{-2}$, (c) $\left(\dfrac{3}{5}\right)^{-2}$.

Answer in your workbook.
ApplyMedium3 MARKS

Q7. Simplify each, writing your answer with positive indices: (a) $a^{-2} \times a^6$, (b) $b^{-3} \times b^{-2}$, (c) $\dfrac{c^4}{c^{-1}}$.

Answer in your workbook.
ReasonHard3 MARKS

Q8. Evaluate $(-3)^{-2}$ and $-3^{-2}$. Show your working and explain why the two answers have different signs.

Answer in your workbook.
Comprehensive Answers

Quick Check

1. A — $\dfrac{1}{9}$.

2. C — $8$.

3. D — $a^2$.

4. A — $\dfrac{9}{4}$.

5. B — $-\dfrac{1}{8}$.

Show Your Working Model Answers

Q6 (3 marks): (a) $5^{-1} = \tfrac{1}{5} = 0.2$ [1]; (b) $\left(\tfrac{1}{4}\right)^{-2} = 4^2 = 16$ [1]; (c) $\left(\tfrac{3}{5}\right)^{-2} = \left(\tfrac{5}{3}\right)^2 = \tfrac{25}{9}$ [1].

Q7 (3 marks): (a) $a^{-2+6} = a^4$ [1]; (b) $b^{-3+(-2)} = b^{-5} = \dfrac{1}{b^5}$ [1]; (c) $c^{4-(-1)} = c^5$ [1].

Q8 (3 marks): $(-3)^{-2} = \dfrac{1}{(-3)^2} = \dfrac{1}{9}$ [1]. $-3^{-2}$ has no bracket, so the negative sign is OUTSIDE: $-3^{-2} = -\dfrac{1}{3^2} = -\dfrac{1}{9}$ [1]. In $(-3)^{-2}$ the base is $-3$, and squaring removes the minus; in $-3^{-2}$ only $3$ is the base, and the minus is applied after [1].

Stretch Challenge · +25 XP, +10 coins

Mixed Indices Marathon

Simplify $\dfrac{a^{-2} \times a^5}{a^{-4}}$ giving your answer with a positive index, then evaluate when $a = 2$.

Reveal solution

Top: $a^{-2+5} = a^3$. Divide: $a^{3-(-4)} = a^7$. With $a = 2$: $2^7 = 128$.

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Quick Review

Flip rule

$\left(\tfrac{a}{b}\right)^{-n} = \left(\tfrac{b}{a}\right)^n$

Combine

$a^{-3} \times a^5 = a^2$

Negative base

$(-2)^{-3} = -\tfrac{1}{8}$

Standard form

Positive indices only

Brackets matter

$(-3)^{-2} = \tfrac{1}{9}$ vs $-3^{-2} = -\tfrac{1}{9}$

Convert at end

$a^{-5} \to \dfrac{1}{a^5}$

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