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Lesson 10 ~25 min Unit 1 · Index Laws +85 XP

Mixed Numerical Index Laws

Combine product, quotient, power-of-a-power, zero and negative indices on numerical bases. Finish every answer with a positive index.

Today's hook: $5^3 \times 5^{-1} \div 5^2$ — three index laws in one expression. Can you crack it in one line?
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Printable Worksheets

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Think First
warm-up

Without using a calculator, predict the value of $\dfrac{2^5 \times 2^{-3}}{2^4}$. What single power of 2 should you get? Convert any negative index to a fraction at the end.

Record in your workbook.
1
The Big Idea
+5 XP

When an expression mixes several index laws on the same base, work it in one column at a time: simplify the numerator, simplify the denominator, then subtract the indices. Apply zero and negative index rules last.

For one base $a$, use $a^m \times a^n = a^{m+n}$, $\dfrac{a^m}{a^n} = a^{m-n}$, $(a^m)^n = a^{mn}$, $a^0 = 1$ and $a^{-n} = \dfrac{1}{a^n}$. Combine indices by adding above the line and subtracting the bottom from the top. Finish with a positive index.

$\dfrac{a^m \times a^p}{a^n} = a^{m+p-n}$
One base, one chain
Same base $\to$ collect indices.
Top minus bottom
$\dfrac{2^5 \times 2^{-3}}{2^4} = 2^{5-3-4} = 2^{-2}$.
Check by evaluation
Always verify with a small base.
2
What You'll Master
objectives

Know

  • All five index laws on a single base
  • Order of combining indices
  • How to write final answers without negative indices

Understand

  • Why same-base expressions collapse to one power
  • Why $a^0 = 1$ acts as “no contribution”
  • Why a negative answer index becomes $\tfrac{1}{a^n}$

Can Do

  • Simplify $\dfrac{2^5 \times 2^{-3}}{2^4}$ to $\tfrac{1}{4}$
  • Simplify $(3^{-2})^3 \times 3^4$ to $\tfrac{1}{9}$
  • Evaluate $\dfrac{5^2 \times 5^0}{5^{-1}}$
3
Words You Need
vocabulary
Mixed expressionUses two or more index laws together.
Same baseRequired to add/subtract indices.
Zero index$a^0 = 1$ (for $a \ne 0$).
Reciprocate$a^{-n} = \dfrac{1}{a^n}$ at the end.
CollapseCombine many powers of one base into a single power.
Check by evaluatingSubstitute and verify with arithmetic.
4
Spot the Trap
heads-up

Wrong: “$\dfrac{2^5 \times 2^{-3}}{2^4} = 2^{5 \times -3 - 4} = 2^{-19}$.” Adding has been replaced with multiplying.

Right: $2^{5+(-3)-4} = 2^{-2} = \dfrac{1}{4}$.

Wrong: “$5^2 \times 5^0 = 0$” — the zero index makes the whole product zero.

Right: $5^0 = 1$, so $5^2 \times 5^0 = 25 \times 1 = 25$.

5
Stacking laws on one base
+5 XP

If every term shares the same base, collect indices into one expression first, then evaluate.

$\dfrac{2^5 \times 2^{-3}}{2^4}$ — top: $5 + (-3) = 2$, so numerator is $2^2$. Then $\dfrac{2^2}{2^4} = 2^{2-4} = 2^{-2}$. Finish: $2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}$.

$\dfrac{2^5 \times 2^{-3}}{2^4} = \dfrac{1}{4}$
6
Power-of-a-power inside the mix
+5 XP

When part of the expression is $(a^m)^n$, multiply the indices before combining with the rest.

$(3^{-2})^3 \times 3^4 = 3^{-2 \times 3} \times 3^4 = 3^{-6} \times 3^4 = 3^{-6+4} = 3^{-2} = \dfrac{1}{9}$. Multiply inside the bracket, then add for the product.

$(3^{-2})^3 \times 3^4 = \dfrac{1}{9}$

Watch Me Solve It · 3 examples

Watch Me Solve It · Product, negative and quotient
+15 XP per step
Q1
PROBLEM
Simplify and evaluate $\dfrac{2^5 \times 2^{-3}}{2^4}$.
  1. 1
    Simplify the numerator
    $2^5 \times 2^{-3} = 2^{5+(-3)} = 2^2$
    Same base — add indices.
  2. 2
    Divide using the quotient rule
    $\dfrac{2^2}{2^4} = 2^{2-4} = 2^{-2}$
  3. 3
    Write with a positive index
    $2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}$
Answer$\dfrac{1}{4}$
Watch Me Solve It · Power then product
+15 XP per step
Q2
PROBLEM
Evaluate $(3^{-2})^3 \times 3^4$, giving a positive answer.
  1. 1
    Power of a power
    $(3^{-2})^3 = 3^{-2 \times 3} = 3^{-6}$
    Multiply the indices.
  2. 2
    Product rule
    $3^{-6} \times 3^4 = 3^{-6+4} = 3^{-2}$
  3. 3
    Positive index
    $3^{-2} = \dfrac{1}{9}$
Answer$\dfrac{1}{9}$
Watch Me Solve It · Zero and negative together
+15 XP per step
Q3
PROBLEM
Evaluate $\dfrac{5^2 \times 5^0}{5^{-1}}$.
  1. 1
    Apply the zero index
    $5^0 = 1 \quad \Rightarrow \quad 5^2 \times 5^0 = 5^2$
  2. 2
    Quotient rule
    $\dfrac{5^2}{5^{-1}} = 5^{2-(-1)} = 5^{3}$
    Subtracting a negative adds.
  3. 3
    Evaluate
    $5^{3} = 125$
Answer$125$

Common Pitfalls

8
Common Pitfalls
heads-up
Multiplying indices when you should add
Product rule is “same base, add indices”. Multiplying is only for $(a^m)^n$.
Fix: Label each step — product (add), quotient (subtract), power (multiply).
Treating $a^0$ as $0$
$5^0$ is $1$, not $0$. It contributes nothing to a multiplication.
Fix: Replace $a^0$ with $1$ as soon as you see it.
Leaving a negative index
$2^{-2}$ is not a final answer.
Fix: Convert to $\dfrac{1}{a^n}$ in the final line.
Copy Into Your Books

Same base, one column

  • $a^m \times a^n = a^{m+n}$
  • $\dfrac{a^m}{a^n} = a^{m-n}$
  • $(a^m)^n = a^{mn}$

Specials

  • $a^0 = 1$
  • $a^{-n} = \dfrac{1}{a^n}$
  • Apply specials last

Strategy

  • Top first
  • Bottom next
  • Top minus bottom

Worked numbers

  • $\dfrac{2^5 \times 2^{-3}}{2^4} = \tfrac{1}{4}$
  • $(3^{-2})^3 \times 3^4 = \tfrac{1}{9}$
  • $\dfrac{5^2 \times 5^0}{5^{-1}} = 125$

How are you completing this lesson?

Brain Trainer · 4 problems

D
Brain Trainer · Mixed numerical drills
4 problems

Combine several index laws on one base each time.

  1. 1 Evaluate $\dfrac{2^7}{2^4 \times 2^{-1}}$.

    Top index 7, bottom $4 + (-1) = 3$. So $2^{7-3}$.$2^4 = 16$

  2. 2 Evaluate $(4^2)^{-1} \times 4^3$.

    $(4^2)^{-1} = 4^{-2}$, then $-2 + 3 = 1$.$4^1 = 4$

  3. 3 Evaluate $\dfrac{3^0 \times 3^5}{3^2}$.

    $3^0 = 1$, top is $3^5$, then $5 - 2 = 3$.$3^3 = 27$

  4. 4 Evaluate $\dfrac{6^{-1} \times 6^4}{6^2}$.

    Top: $-1 + 4 = 3$. Then $3 - 2 = 1$.$6^1 = 6$
Complete in your workbook.

Quick Check · 5 questions

1
Evaluate $\dfrac{2^5 \times 2^{-3}}{2^4}$.
+10 XP
2
Evaluate $(3^{-2})^3 \times 3^4$.
+10 XP
3
Evaluate $\dfrac{5^2 \times 5^0}{5^{-1}}$.
+10 XP
4
Evaluate $\dfrac{2^7}{2^4 \times 2^{-1}}$.
+10 XP
5
Simplify $7^{-2} \times 7^2$.
+10 XP

Show Your Working · 3 questions

Show Your Working
9 marks total
ApplyMedium3 MARKS

Q6. Evaluate each, showing index working before any arithmetic: (a) $\dfrac{2^6 \times 2^{-2}}{2^3}$, (b) $4^3 \times 4^{-3}$, (c) $\dfrac{10^0 \times 10^4}{10^2}$.

Answer in your workbook.
ApplyMedium3 MARKS

Q7. Simplify, writing your answer as a positive whole number or fraction: (a) $(2^3)^{-1} \times 2^5$, (b) $\dfrac{3^{-2} \times 3^4}{3}$, (c) $\dfrac{6^2 \times 6^{-3}}{6^{-2}}$.

Answer in your workbook.
ReasonHard3 MARKS

Q8. A student wrote “$\dfrac{2^3 \times 2^{-5}}{2^{-1}} = 2^{3 - 5 - 1} = 2^{-3} = \dfrac{1}{8}$.” Identify the error, fix the working, and give the correct answer with a positive index.

Answer in your workbook.
Comprehensive Answers

Quick Check

1. B — $\dfrac{1}{4}$.

2. C — $\dfrac{1}{9}$.

3. D — $125$.

4. A — $16$.

5. C — $1$.

Show Your Working Model Answers

Q6 (3 marks): (a) $\dfrac{2^{6+(-2)}}{2^3} = \dfrac{2^4}{2^3} = 2^1 = 2$ [1]; (b) $4^{3+(-3)} = 4^0 = 1$ [1]; (c) $\dfrac{1 \times 10^4}{10^2} = 10^{4-2} = 10^2 = 100$ [1].

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

Q8 (3 marks): Error: when dividing by $2^{-1}$ you must SUBTRACT $-1$, which adds $1$. The student subtracted $1$ instead [1]. Correct working: $\dfrac{2^3 \times 2^{-5}}{2^{-1}} = 2^{3 + (-5) - (-1)} = 2^{3 - 5 + 1} = 2^{-1}$ [1]. So the answer is $2^{-1} = \dfrac{1}{2}$ [1].

Stretch Challenge · +25 XP, +10 coins

Five Laws, One Line

Evaluate $\dfrac{(2^3)^2 \times 2^{-4}}{2^0 \times 2^{-1}}$, giving your final answer with a positive index. Show every step.

Reveal solution

$(2^3)^2 = 2^6$. Top: $2^{6 + (-4)} = 2^2$. Bottom: $2^{0 + (-1)} = 2^{-1}$. So $\dfrac{2^2}{2^{-1}} = 2^{2 - (-1)} = 2^3 = 8$.

R
Quick Review

Product rule

$a^m \times a^n = a^{m+n}$

Quotient rule

$\dfrac{a^m}{a^n} = a^{m-n}$

Power rule

$(a^m)^n = a^{mn}$

Zero index

$a^0 = 1$

Negative index

$a^{-n} = \dfrac{1}{a^n}$

Final form

Positive indices only

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