Mathematics • Year 9 • Unit 1 • Lesson 10
Mixed Numerical Index Laws
Build fluency combining all five index rules on numerical bases: product (add), quotient (subtract), power (multiply), zero ($=1$) and negative (reciprocate). Finish every answer with a positive index.
1. I do — fully worked example
Read every step. Each line has a short reason on the right.
Problem. Simplify and evaluate $\dfrac{2^5 \times 2^{-3}}{2^4}$.
Step 1 — Simplify the numerator (product rule).
$2^5 \times 2^{-3} = 2^{5 + (-3)} = 2^2$
Reason: same base $2$ on both sides of the $\times$ — ADD the indices, even when one is negative.
Step 2 — Divide using the quotient rule.
$\dfrac{2^2}{2^4} = 2^{2-4} = 2^{-2}$
Reason: same base divided — SUBTRACT (top minus bottom).
Step 3 — Write with a positive index.
$2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}$
Reason: standard form needs positive indices only — reciprocate to remove the negative.
Answer: $\mathbf{\dfrac{1}{4}}$.
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 $(3^{-2})^3 \times 3^4$, giving a positive answer with a positive index.
Step 1 — Power-of-a-power on the bracket: __________________ the indices inside.
$(3^{-2})^3 = 3^{(\,-2\,) \times (\,\_\_\,)} = 3^{\_\_\_}$
Step 2 — Product rule on the same base: __________ the indices.
$3^{-6} \times 3^4 = 3^{(\,-6\,) + (\,\_\_\,)} = 3^{\_\_\_}$
Step 3 — Convert to positive index:
$3^{-2} = \dfrac{1}{3^{\_\_\,}} = \dfrac{1}{\_\_\_}$
Step 4 — Final answer:
$(3^{-2})^3 \times 3^4 = \_\_\_\_\_\_$
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (two rules). The middle two are standard (three rules including zero or negative). The last two are extension (four+ rules in one expression).
Foundation — two rules combined
3.1 Evaluate $\dfrac{2^7}{2^4 \times 2^{-1}}$. 1 mark
3.2 Evaluate $(4^2)^{-1} \times 4^3$. 1 mark
3.3 Simplify $7^{-2} \times 7^2$. 1 mark
3.4 Evaluate $4^3 \times 4^{-3}$. 1 mark
Standard — three rules including zero or negative
3.5 Evaluate $\dfrac{3^0 \times 3^5}{3^2}$. 2 marks
3.6 Evaluate $\dfrac{5^2 \times 5^0}{5^{-1}}$. 2 marks
Extension — four+ rules in one expression
3.7 Evaluate $\dfrac{(2^3)^2 \times 2^{-4}}{2^0 \times 2^{-1}}$. Give your final answer with a positive index. 3 marks
3.8 Evaluate $\dfrac{6^{-1} \times 6^4}{6^2}$. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $(3^{-2})^3 \times 3^4$)
Step 1: multiply the indices in the bracket. $(3^{-2})^3 = 3^{(-2) \times \mathbf{3}} = 3^{\mathbf{-6}}$.
Step 2: add the indices. $3^{-6} \times 3^4 = 3^{(-6) + \mathbf{4}} = 3^{\mathbf{-2}}$.
Step 3: $3^{-2} = \dfrac{1}{3^{\mathbf{2}}} = \dfrac{1}{\mathbf{9}}$.
Step 4: $(3^{-2})^3 \times 3^4 = \mathbf{\dfrac{1}{9}}$.
3.1 — $\dfrac{2^7}{2^4 \times 2^{-1}}$
Bottom: $2^{4 + (-1)} = 2^3$. Then $\dfrac{2^7}{2^3} = 2^{7-3} = 2^4 = \mathbf{16}$.
3.2 — $(4^2)^{-1} \times 4^3$
$(4^2)^{-1} = 4^{2 \times (-1)} = 4^{-2}$. Then $4^{-2} \times 4^3 = 4^{-2+3} = 4^1 = \mathbf{4}$.
3.3 — $7^{-2} \times 7^2$
Product rule: $7^{-2+2} = 7^0 = \mathbf{1}$ (zero-index rule).
3.4 — $4^3 \times 4^{-3}$
Product rule: $4^{3 + (-3)} = 4^0 = \mathbf{1}$.
3.5 — $\dfrac{3^0 \times 3^5}{3^2}$
$3^0 = 1$, so numerator is $1 \times 3^5 = 3^5$. Quotient: $\dfrac{3^5}{3^2} = 3^{5-2} = 3^3 = \mathbf{27}$.
3.6 — $\dfrac{5^2 \times 5^0}{5^{-1}}$
$5^0 = 1$, so numerator is $5^2$. Quotient: $\dfrac{5^2}{5^{-1}} = 5^{2 - (-1)} = 5^{2+1} = 5^3 = \mathbf{125}$.
3.7 — $\dfrac{(2^3)^2 \times 2^{-4}}{2^0 \times 2^{-1}}$
Power: $(2^3)^2 = 2^6$. Numerator: $2^6 \times 2^{-4} = 2^{6+(-4)} = 2^2$.
Denominator: $2^0 \times 2^{-1} = 2^{0+(-1)} = 2^{-1}$.
Whole expression: $\dfrac{2^2}{2^{-1}} = 2^{2-(-1)} = 2^3 = \mathbf{8}$.
3.8 — $\dfrac{6^{-1} \times 6^4}{6^2}$
Top: $6^{-1+4} = 6^3$. Whole: $\dfrac{6^3}{6^2} = 6^{3-2} = 6^1 = \mathbf{6}$.