Mathematics • Year 9 • Unit 1 • Lesson 9
Evaluating Negative Indices in the Real World
Use the flip rule for fractions, combine negative + positive indices, and convert to standard form (positive indices only) in everyday settings: gaming damage multipliers, scaled photos, audio loudness, recipe halving, and tournament brackets.
1. Word problems
Each problem uses rules from Lesson 9: $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$, combine indices using product/quotient rules even when negative, and finish with positive indices. Show your working.
1.1 — Damage multiplier in a game. A boss fight has a damage multiplier of $\left(\dfrac{1}{2}\right)^{-3}$ on a particular attack. The game designers want to know how many times the base damage that is.
(a) Evaluate $\left(\dfrac{1}{2}\right)^{-3}$ using the flip rule.
(b) Write a one-sentence interpretation in plain English (e.g. "the attack does $\ldots$ times the base damage"). 3 marks
1.2 — Recipe stretched the wrong way. A baker's recipe is supposed to be scaled by $\left(\dfrac{3}{2}\right)^2$ to feed a bigger group. By mistake, they scale it by $\left(\dfrac{3}{2}\right)^{-2}$ instead.
(a) Evaluate both multipliers and state which gives more food.
(b) By what factor (as a fraction) does the mistaken multiplier shrink the recipe compared with the intended one? 3 marks
1.3 — Sound getting quieter. Every time you halve the volume, the sound power drops by a factor of $\dfrac{1}{2}$. After $n$ halvings the power is $\left(\dfrac{1}{2}\right)^n$ of the original — equivalently, $2^{-n}$ times the original.
(a) Show that $\left(\dfrac{1}{2}\right)^n$ and $2^{-n}$ are the same thing using the flip rule.
(b) After $5$ halvings, what fraction of the original power is left? Use $2^{-5}$ and convert to a fraction. 3 marks
1.4 — Two halving streams combined. A streamer's number of live viewers follows a pattern: it halves $2$ times during a sleep break, then jumps up by a factor of $2^5$ when they go live again. The combined multiplier is $\dfrac{1}{2^2} \times 2^5 = 2^{-2} \times 2^5$.
(a) Use the product rule to combine $2^{-2} \times 2^5$ into a single power of $2$.
(b) Evaluate that single power as a number. Interpret what it means for the viewer count. 3 marks
1.5 — Mixed multiplier (compare to one round). A round of a card-game tournament has a "score factor" of $\dfrac{a^{-2} \times a^5}{a^{-4}}$, where $a$ is the player's current rating.
(a) Simplify the score factor to a single power of $a$ with a positive index.
(b) Evaluate the score factor when $a = 2$. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate writes "$(-3)^{-2} = -\dfrac{1}{9}$, because the base is negative". In your own words, explain (i) how the negative-index rule should be applied to $(-3)^{-2}$, (ii) why squaring a negative number gives a positive result, and (iii) what the correct value is, with one sentence on how it differs from $-3^{-2}$ (no brackets).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Damage multiplier
(a) Flip: $\left(\dfrac{1}{2}\right)^{-3} = 2^3 = \mathbf{8}$.
(b) The attack does $\mathbf{8}$ times the base damage.
1.2 — Recipe stretched the wrong way
(a) Intended: $\left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4} = 2.25$. Mistake: $\left(\dfrac{3}{2}\right)^{-2} = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9} \approx 0.44$. The intended multiplier gives MORE food (more than double); the mistake gives less than half.
(b) Ratio: $\dfrac{(3/2)^{-2}}{(3/2)^2} = (3/2)^{-2-2} = (3/2)^{-4} = (2/3)^4 = \dfrac{16}{81}$. So the mistaken recipe is only $\mathbf{\dfrac{16}{81}}$ of the intended size — roughly one-fifth.
1.3 — Sound getting quieter
(a) $\left(\dfrac{1}{2}\right)^n = \dfrac{1^n}{2^n} = \dfrac{1}{2^n} = 2^{-n}$. ✓ (same thing).
(b) After $5$ halvings: $2^{-5} = \dfrac{1}{2^5} = \mathbf{\dfrac{1}{32}}$ of the original power.
1.4 — Two halving streams combined
(a) Product rule (negative indices OK): $2^{-2} \times 2^5 = 2^{-2+5} = \mathbf{2^3}$.
(b) $2^3 = \mathbf{8}$. The viewer count ends up $8$ times higher than it was BEFORE the sleep break.
1.5 — Mixed multiplier
(a) Numerator: $a^{-2} \times a^5 = a^{-2+5} = a^3$. Then $\dfrac{a^3}{a^{-4}} = a^{3 - (-4)} = a^{3+4} = \mathbf{a^7}$.
(b) With $a = 2$: $2^7 = \mathbf{128}$.
2.1 — Explain your thinking (sample response)
To apply the negative-index rule to $(-3)^{-2}$, treat the WHOLE base $-3$ as the "$a$". So $(-3)^{-2} = \dfrac{1}{(-3)^2}$. Now $(-3)^2$ means $-3$ multiplied by itself: $(-3) \times (-3) = +9$ (two negatives multiply to a positive). So $(-3)^{-2} = \dfrac{1}{9} = \mathbf{\dfrac{1}{9}}$, a positive fraction.
The classmate's answer "$-\dfrac{1}{9}$" is wrong: the negative sign on the base disappears when we SQUARE it (even index).
Compare with $-3^{-2}$ (no brackets): now the base is only $3$, and the minus sits outside. $-3^{-2} = -(3^{-2}) = -\dfrac{1}{9}$. So $(-3)^{-2}$ and $-3^{-2}$ disagree because of where the brackets are — brackets decide whether the minus is part of the base or not.
Marking: 1 mark for correctly applying the rule with the WHOLE bracket as base; 1 for showing $(-3)^2 = +9$; 1 for the correct answer $\dfrac{1}{9}$; 1 for the bracket-difference comparison.