Combining Index Laws
When two or three rules collide in one expression: simplify inside brackets first, multiply next, divide last.
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Simplify $2^3 \times 2^5 \div 2^4$. Which rule do you use first, the product rule or the quotient rule? Does the order change your answer?
Real expressions usually need more than one index law. The trick is choosing a sensible order so you simplify, not complicate.
Strategy: deal with brackets first (power-of-a-power), then multiply (product rule), then divide (quotient rule). Keep the same base.
Know
- Product rule: $a^m \times a^n = a^{m+n}$
- Quotient rule: $a^m \div a^n = a^{m-n}$
- Power of a power: $(a^m)^n = a^{mn}$
Understand
- Why brackets are simplified first
- Why order changes intermediate steps but not the final answer (if rules used correctly)
- How to check by evaluating numerically
Can Do
- Simplify $2^3 \times 2^5 \div 2^4$
- Simplify $(3^2)^3 \times 3^4$
- Verify answers by direct calculation
Wrong: "$(3^2)^3 \times 3^4 = 3^2 \times 3^{3+4}$", applying the product rule before resolving the bracket.
Right: Resolve bracket first: $(3^2)^3 = 3^6$, then $3^6 \times 3^4 = 3^{10}$.
Wrong: "$2^3 \times 2^5 \div 2^4 = 2^{3 \times 5 - 4} = 2^{11}$", multiplying indices instead of adding.
Right: $2^{3+5-4} = 2^4 = 16$. Product = add; quotient = subtract.
Whenever an expression mixes rules, follow this checklist:
1. Brackets apply $(a^m)^n = a^{mn}$ first.
2. Multiply combine factors with the same base: $a^m \times a^n = a^{m+n}$.
3. Divide finish with $a^m \div a^n = a^{m-n}$.
4. Check if the numbers are small, evaluate directly.
When the base is small (like $2$ or $3$), you can re-do the problem by direct calculation as a check.
For $2^3 \times 2^5 \div 2^4$: by index laws we get $2^4 = 16$. Verify directly: $8 \times 32 \div 16 = 256 \div 16 = 16$. The two methods agree, we know we got it right.
Watch Me Solve It · 3 examples
- 1Product rule first$2^3 \times 2^5 = 2^{3+5} = 2^8$Same base, indices add.
- 2Quotient rule$2^8 \div 2^4 = 2^{8-4} = 2^4$
- 3Evaluate & check$2^4 = 16$Verify: $8 \times 32 \div 16 = 16$ ✓
- 1Resolve the bracket$(3^2)^3 = 3^{2 \times 3} = 3^6$Power-of-a-power: multiply indices.
- 2Apply product rule$3^6 \times 3^4 = 3^{6+4}$
- 3Final index$3^{10}$
- 1Resolve the bracket$(5^3)^2 = 5^{3 \times 2} = 5^6$
- 2Apply quotient rule$\dfrac{5^6}{5^4} = 5^{6-4} = 5^2$
- 3Evaluate$5^2 = 25$Verify: $\dfrac{15625}{625} = 25$ ✓
Common Pitfalls
Strategy
- 1. Brackets first
- 2. Multiply (product rule)
- 3. Divide (quotient rule)
- 4. Check with numbers
Rules summary
- $a^m \times a^n = a^{m+n}$
- $a^m \div a^n = a^{m-n}$
- $(a^m)^n = a^{mn}$
Example
- $(3^2)^3 \times 3^4$
- $= 3^6 \times 3^4$
- $= 3^{10}$
Always check
- Same base?
- Bracket resolved first?
- Add or multiply?
How are you completing this lesson?
Brain Trainer · 4 problems
Four drills to lock in the strategy.
1 Simplify $2^4 \times 2^2 \div 2^3$.
Add then subtract.$2^3 = 8$2 Simplify $(a^3)^2 \times a^5$.
Bracket: $a^6$, then add 5.$a^{11}$3 Simplify $\dfrac{(2^4)^2}{2^5}$.
$2^8 \div 2^5$.$2^3 = 8$4 Simplify $x^2 \times (x^3)^2 \div x^4$.
$x^2 \times x^6 \div x^4 = x^{2+6-4}$.$x^4$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Simplify each: (a) $4^3 \times 4^2 \div 4^4$, (b) $(a^2)^4 \div a^3$, (c) $(2^3)^2 \times 2^4$.
Q7. A student writes $(a^2)^3 \times a^4 = a^{2+3+4} = a^9$. Identify the error and show the correct working leading to $a^{10}$.
Q8. Simplify $\dfrac{(2^4)^2 \times 2^3}{2^6}$ and then verify your answer by direct calculation.
Quick Check
1. C$2^4 = 16$.
2. D$3^{10}$.
3. A$25$.
4. B$x$.
5. C$8$.
Show Your Working Model Answers
Q6 (3 marks): (a) $4^{3+2-4} = 4^1 = 4$ [1]; (b) $a^{8-3} = a^5$ [1]; (c) $2^6 \times 2^4 = 2^{10}$ [1].
Q7 (2 marks): The student added all three indices, but the bracket $(a^2)^3$ requires MULTIPLYING: $(a^2)^3 = a^6$ [1]. Then product rule: $a^6 \times a^4 = a^{10}$ [1].
Q8 (4 marks): $(2^4)^2 = 2^8$ [1]. Numerator: $2^8 \times 2^3 = 2^{11}$ [1]. Divide: $2^{11} \div 2^6 = 2^5 = 32$ [1]. Verify: $\dfrac{256 \times 8}{64} = \dfrac{2048}{64} = 32$ ✓ [1].
Mixed Bag
Simplify $\dfrac{(a^3)^4 \times a^2}{(a^5)^2}$ giving your answer as a single power of $a$.
Reveal solution
Top: $a^{12} \times a^2 = a^{14}$. Bottom: $a^{10}$. So $a^{14-10} = a^4$.
Brackets first
$(a^m)^n = a^{mn}$
Product
$a^m \times a^n = a^{m+n}$
Quotient
$a^m \div a^n = a^{m-n}$
Strategy
B $\to$ M $\to$ D $\to$ Check
Same base
Required for product/quotient
Verify
Re-do with numbers when small
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