Mathematics • Year 9 • Unit 1 • Lesson 2
Evaluating Powers
Build fluency with evaluating powers of positive, negative and fractional bases — including the crucial difference between $-2^4$ and $(-2)^4$, and BIDMAS with indices. One step at a time, from a fully worked example through guided practice to independent problems.
1. I do — fully worked example
Read every line. Each step has a short reason on the right.
Problem. Evaluate $(-2)^4$ and $-2^4$ separately, then explain why they differ.
Step 1 — Identify the base in each expression.
In $(-2)^4$, the brackets contain the base: base $= -2$.
In $-2^4$, only the $2$ is the base; the negative sits OUTSIDE.
Reason: the index only applies to whatever is directly inside the brackets (or the symbol immediately to the left if there are no brackets).
Step 2 — Evaluate $(-2)^4$.
$(-2)^4 = (-2) \times (-2) \times (-2) \times (-2)$
$= 4 \times 4 = 16$
Reason: even index of a negative base $\to$ positive answer (every pair of negatives multiplies to a positive).
Step 3 — Evaluate $-2^4$.
$-2^4 = -(2 \times 2 \times 2 \times 2) = -(16) = -16$
Reason: the base is just $2$, so $2^4 = 16$. The negative sign applies LAST.
Step 4 — Compare.
$(-2)^4 = 16$ but $-2^4 = -16$. Brackets change the base.
Answer: $(-2)^4 = \mathbf{16}$ and $-2^4 = \mathbf{-16}$.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. Evaluate $\left(\dfrac{2}{3}\right)^3$.
Step 1 — Spot the rule: for a power of a fraction, raise the __________ and the __________ separately.
Step 2 — Apply $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$:
$\left(\dfrac{2}{3}\right)^3 = \dfrac{\_\_^3}{\_\_^3}$
Step 3 — Evaluate the top:
$2^3 = 2 \times 2 \times 2 = \_\_\_\_\_$
Step 4 — Evaluate the bottom:
$3^3 = 3 \times 3 \times 3 = \_\_\_\_\_$
Step 5 — Combine:
$\left(\dfrac{2}{3}\right)^3 = \dfrac{\_\_\_}{\_\_\_}$
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (single skill). The middle two are standard (combine two ideas). The last two are extension (push your thinking).
Foundation — single skill
3.1 Evaluate $(-3)^2$. 1 mark
3.2 Evaluate $-3^2$. 1 mark
3.3 Evaluate $\left(\dfrac{1}{4}\right)^2$. 1 mark
3.4 Evaluate $1^{50}$. 1 mark
Standard — combine two ideas
3.5 Evaluate $(-2)^5$. 2 marks
3.6 Evaluate $\;5 + 3 \times 2^3\;$ using BIDMAS. Show every step. 2 marks
Extension — push your thinking
3.7 Evaluate $\;(-1)^{99} + (-1)^{100}\;$. Use the even/odd index rule. 3 marks
3.8 Evaluate $\;20 - 2 \times (-3)^2\;$. Show every step of BIDMAS. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $(2/3)^3$)
Step 1: raise the numerator (top) and denominator (bottom) separately.
Step 2: $\left(\dfrac{2}{3}\right)^3 = \dfrac{\mathbf{2}^3}{\mathbf{3}^3}$.
Step 3: $2^3 = \mathbf{8}$.
Step 4: $3^3 = \mathbf{27}$.
Step 5: $\left(\dfrac{2}{3}\right)^3 = \mathbf{\dfrac{8}{27}}$.
3.1 — $(-3)^2$
Base is $-3$ (inside brackets). Even index $\to$ positive: $(-3)^2 = (-3) \times (-3) = \mathbf{9}$.
3.2 — $-3^2$
No brackets — base is just $3$. So $-3^2 = -(3 \times 3) = -(9) = \mathbf{-9}$.
Compare with 3.1: brackets make all the difference.
3.3 — $(1/4)^2$
Square top and bottom: $\dfrac{1^2}{4^2} = \mathbf{\dfrac{1}{16}}$.
3.4 — $1^{50}$
$1$ raised to ANY power is $1$, since $1 \times 1 \times \ldots \times 1 = 1$. So $1^{50} = \mathbf{1}$.
3.5 — $(-2)^5$
Base $-2$, odd index $\to$ negative answer.
$(-2)^5 = (-2)(-2)(-2)(-2)(-2) = (4)(4)(-2) = 16 \times (-2) = \mathbf{-32}$.
3.6 — $5 + 3 \times 2^3$
BIDMAS — index first:
$2^3 = 8$.
Multiplication next: $3 \times 8 = 24$.
Addition last: $5 + 24 = \mathbf{29}$.
3.7 — $(-1)^{99} + (-1)^{100}$
$99$ is odd, so $(-1)^{99} = -1$.
$100$ is even, so $(-1)^{100} = +1$.
Sum: $-1 + 1 = \mathbf{0}$.
3.8 — $20 - 2 \times (-3)^2$
BIDMAS — index first: $(-3)^2 = 9$ (brackets, so base is $-3$, even index).
Multiplication: $2 \times 9 = 18$.
Subtraction: $20 - 18 = \mathbf{2}$.