00-11 — Laws of Indices (Exponents)

Phase: 0 — Arithmetic & Number Foundations Subject: 00-11 Prerequisites: 00-06 — Powers and Roots Next subject: 00-12 — Scientific Notation and Standard Form

Learning Objectives

By the end of this subject, you will be able to:

1. State, derive, and apply all five fundamental laws of indices (product, quotient, power of power, power of product, power of quotient) to simplify complex expressions
2. Extend the laws to zero, negative, and fractional indices with both numerical and algebraic expressions
3. Solve equations where the unknown is in the exponent by equating bases
4. Simplify expressions involving mixed indices, including multi-step simplifications that combine several laws
5. Recognize common index-notation patterns and pitfalls, including why (a+b)ⁿ ≠ aⁿ + bⁿ

Core Content

1. Review — What Is an Index (Exponent)?

The index (plural: indices) or exponent tells you how many times the base is multiplied by itself:

aⁿ = a × a × a × ... × a (n factors of a)

Where: - a is the base - n is the index (or exponent or power)

In Subject 00-06, you learned the five fundamental laws. Here we go deeper: we prove each law, apply them to algebraic expressions with variables, and handle edge cases.

⚠️ THIS IS CRITICAL — The laws of indices are the foundation for algebraic manipulation of powers. They appear in every branch of mathematics: polynomial algebra, logarithms, exponential growth/decay, scientific notation, calculus, and probability. Mastering them now saves enormous effort later.

2. Law 1 — Product of Powers: aᵐ × aⁿ = a^(m+n)

Statement: When multiplying powers with the same base, ADD the indices.

Proof (for positive integer indices):

$aᵐ × aⁿ = (a × a × ... × a) × (a × a × ... × a)
          \_____ m factors _____/   \_____ n factors _____/
        = a × a × ... × a   (m + n factors total)
        = a^(m+n)
$

Examples: - 7³ × 7⁴ = 7^(3+4) = 7⁷ = 823,543 - x⁵ × x² = x⁷ - 3² × 3³ × 3 = 3^(2+3+1) = 3⁶ = 729 (remember 3 = 3¹) - 2ᵃ × 2ᵇ = 2^(a+b)

Key insight: The bases MUST be the same. 2³ × 3⁴ cannot be simplified using this law.

3. Law 2 — Quotient of Powers: aᵐ ÷ aⁿ = a^(m−n) (a ≠ 0)

Statement: When dividing powers with the same base, SUBTRACT the indices.

Proof:

aᵐ     a × a × ... × a (m factors)
--- = ------------------------------
aⁿ     a × a × ... × a (n factors)

Cancel n factors from numerator and denominator,
leaving m − n factors in the numerator:

= a × a × ... × a (m−n factors) = a^(m−n)

Examples: - 5⁸ ÷ 5³ = 5^(8−3) = 5⁵ = 3,125 - y⁷ ÷ y² = y⁵ - 10¹² ÷ 10⁴ = 10⁸ = 100,000,000 - aᵖ ÷ aᑫ = a^(p−q)

What if m = n? Then aᵐ ÷ aᵐ = a^(m−m) = a⁰ = 1 (which makes sense — any non-zero number divided by itself is 1).

What if m < n? Then we get a negative index — see Section 6.

4. Law 3 — Power of a Power: (aᵐ)ⁿ = a^(m×n)

Statement: When raising a power to another power, MULTIPLY the indices.

Proof:

$(aᵐ)ⁿ = aᵐ × aᵐ × ... × aᵐ    (n copies of aᵐ)
       = a^(m + m + ... + m)   (n times, by Law 1)
       = a^(m × n)
$

Examples: - (2³)⁴ = 2^(3×4) = 2¹² = 4,096 - (x²)⁵ = x¹⁰ - (10²)³ = 10⁶ = 1,000,000

⚠️ THIS IS CRITICAL — (aᵐ)ⁿ is NOT the same as a^(mⁿ). - (2³)⁴ = 2¹² = 4,096 - 2^(3⁴) = 2⁸¹ ≈ 2.4 × 10²⁴ (enormously different!)

The parentheses matter enormously. (aᵐ)ⁿ means multiply the indices; a^(mⁿ) means raise m to the nth power first.

5. Law 4 — Power of a Product: (a × b)ⁿ = aⁿ × bⁿ

Statement: A power distributes over multiplication.

Proof:

$(a × b)ⁿ = (a × b) × (a × b) × ... × (a × b)   (n factors)
         = (a × a × ... × a) × (b × b × ... × b) (rearranging — multiplication commutes)
         = aⁿ × bⁿ
$

Examples: - (2 × 5)³ = 2³ × 5³ = 8 × 125 = 1,000 - (xy)⁴ = x⁴y⁴ - (3p)² = 3² × p² = 9p² - (−2 × 3)⁵ = (−2)⁵ × 3⁵ = (−32) × 243 = −7,776

⚠️ THIS IS CRITICAL — (a + b)ⁿ does NOT equal aⁿ + bⁿ. - (3 + 2)² = 5² = 25 - But 3² + 2² = 9 + 4 = 13 - The law only works for multiplication, not addition!

The distributive property of exponentiation is limited to multiplication and division. This is one of the most common algebraic errors.

6. Law 5 — Power of a Quotient: (a / b)ⁿ = aⁿ / bⁿ (b ≠ 0)

Statement: A power distributes over division.

Proof: Same reasoning as Law 4.

$(a / b)ⁿ = (a/b) × (a/b) × ... × (a/b) = aⁿ / bⁿ
$

Examples: - (2/3)⁴ = 2⁴ / 3⁴ = 16/81 - (x/y)³ = x³ / y³ - (4/5)² = 16/25 - (a/b)ⁿ = aⁿ / bⁿ (this holds for any real a, b with b ≠ 0)

7. Zero Index: a⁰ = 1 (a ≠ 0)

Derivation from Law 2:

Take a³ ÷ a³. By Law 2: a³ ÷ a³ = a^(3−3) = a⁰. But a³ ÷ a³ = 1 (any non-zero number divided by itself is 1). Therefore a⁰ = 1.

This holds for ANY non-zero base: - 5⁰ = 1 - (−3)⁰ = 1 - (1,000,000)⁰ = 1 - (x)⁰ = 1 (for x ≠ 0)

0⁰ is undefined — it is what mathematicians call an "indeterminate form." You cannot consistently assign it a value. We treat it as undefined; you'll learn why in Calculus (Phase 4).

8. Negative Indices: a^(−n) = 1 / aⁿ (a ≠ 0)

Derivation from Law 2:

Take a⁰ ÷ aⁿ. By Law 2: a⁰ ÷ aⁿ = a^(0−n) = a^(−n). But a⁰ = 1, so 1 ÷ aⁿ = 1/aⁿ. Therefore a^(−n) = 1/aⁿ.

Examples: - 2^(−3) = 1/2³ = 1/8 - 10^(−1) = 1/10 = 0.1 - 10^(−2) = 1/100 = 0.01 - x^(−4) = 1/x⁴ - (1/3)^(−2) = 1 / (1/3)² = 1 / (1/9) = 9 (alternatively: (1/3)^(−2) = 3² = 9)

Key insight: A negative index means "take the reciprocal." Moving a factor from numerator to denominator (or vice versa) flips the sign of its index:

$a^(−n)    1           1
------ = -----   and  ------ = aⁿ
  1       aⁿ         a^(−n)
$

Two important conventions:

1. Answers are usually written with positive indices: write x^(−3) as 1/x³.
2. When simplifying, collect all factors with the same base.

Example: Simplify (2x³y^(−2)) × (3x^(−1)y⁴)

$= 2 × 3 × x³ × x^(−1) × y^(−2) × y⁴
= 6 × x^(3+(−1)) × y^((−2)+4)
= 6 × x² × y²
= 6x²y²
$

9. Fractional Indices: a^(1/n) = ⁿ√a

Definition: a^(1/n) is the nth root of a.

a^(1/n) = ⁿ√a, meaning (a^(1/n))ⁿ = a

Why this works with the laws: (a^(1/n))ⁿ = a^((1/n) × n) = a¹ = a. So a^(1/n) satisfies the definition of the nth root.

Examples: - 9^(1/2) = √9 = 3 - 8^(1/3) = ∛8 = 2 - 16^(1/4) = 2 (since 2⁴ = 16) - 32^(1/5) = 2 (since 2⁵ = 32) - 10,000^(1/4) = 10 (since 10⁴ = 10,000)

Negative fractional indices are possible: - 8^(−1/3) = 1 / 8^(1/3) = 1/2 - 64^(−1/2) = 1 / √64 = 1/8

General fractional index: a^(m/n) = (ⁿ√a)ᵐ = ⁿ√(aᵐ)

This combines Law 3 with the nth root: a^(m/n) = a^((1/n) × m) = (a^(1/n))ᵐ = (ⁿ√a)ᵐ Alternatively: a^(m/n) = a^(m × (1/n)) = (aᵐ)^(1/n) = ⁿ√(aᵐ)

Examples: - 8^(2/3) = (∛8)² = 2² = 4 - 16^(3/4) = (⁴√16)³ = 2³ = 8 - 27^(2/3) = (∛27)² = 3² = 9 - 32^(3/5) = (⁵√32)³ = 2³ = 8

Strategy: Usually taking the root first then the power gives smaller numbers and is easier.

10. Solving Equations with Indices

When the unknown is in the exponent, you can often solve by making the bases the same.

Principle: If aᵐ = aⁿ and a > 0, a ≠ 1, then m = n.

Example 1: Solve 2ˣ = 32

$2ˣ = 2⁵
∴ x = 5
$

Example 2: Solve 3^(2x+1) = 27

$3^(2x+1) = 3³
∴ 2x + 1 = 3
   2x = 2
   x = 1
$

Example 3: Solve 4ˣ = 8^(x−1)

(2²)ˣ = (2³)^(x−1)         ← Express both bases as powers of 2
2^(2x) = 2^(3x−3)
∴ 2x = 3x − 3
   x = 3

Example 4: Solve 9ˣ × 3^(x+1) = 81

(3²)ˣ × 3^(x+1) = 3⁴       ← Express all as powers of 3
3^(2x) × 3^(x+1) = 3⁴
3^(2x+x+1) = 3⁴             ← Law 1
3^(3x+1) = 3⁴
∴ 3x + 1 = 4
   3x = 3
   x = 1

11. Simplifying Complex Index Expressions

Strategy for multi-step simplification:

1. Apply Law 3 (power of a power) to remove nested exponents: (aᵐ)ⁿ → a^(mn)
2. Apply Law 4 and Law 5 to distribute exponents over products and quotients
3. Group terms with the same base
4. Apply Law 1 (add indices for multiplication) and Law 2 (subtract indices for division)
5. Convert negative indices to positive in the final answer

Example 1: Simplify (x²y³)⁴ ÷ (x³y²)²

= (x⁸y¹²) ÷ (x⁶y⁴)          ← Law 3 and Law 4
= x^(8−6) × y^(12−4)        ← Law 2 for each base
= x²y⁸

Example 2: Simplify (2a³b^(−2))² × (a²b³)^(−1)

= 2² × a⁶ × b^(−4) × a^(−2) × b^(−3)     ← Law 4 and Law 3, negative index for second factor
= 4 × a^(6+(−2)) × b^((−4)+(−3))          ← Group same bases
= 4 × a⁴ × b^(−7)
= 4a⁴ / b⁷                                 ← Positive indices

Example 3: Simplify (27x⁶y³)^(1/3)

$= 27^(1/3) × x^(6×1/3) × y^(3×1/3)       ← Law 4 and Law 3
= 3 × x² × y¹                             ← ∛27 = 3, 6/3 = 2, 3/3 = 1
= 3x²y
$

12. Summary Table of All Index Laws

Worked Examples

Example 1: Multi-Step Simplification

Simplify: (3x²y³)³ × (2xy)² / (6x⁵y³)²

Solution:

$(3x²y³)³ × (2xy)² / (6x⁵y³)²

Numerator:
(3x²y³)³ = 3³ × (x²)³ × (y³)³ = 27 × x⁶ × y⁹ = 27x⁶y⁹
(2xy)² = 2² × x² × y² = 4x²y²
Product: 27x⁶y⁹ × 4x²y² = 108 × x⁶⁺² × y⁹⁺² = 108x⁸y¹¹

Denominator:
(6x⁵y³)² = 6² × (x⁵)² × (y³)² = 36 × x¹⁰ × y⁶ = 36x¹⁰y⁶

Division:
108x⁸y¹¹ / 36x¹⁰y⁶ = (108/36) × x^(8−10) × y^(11−6)
= 3 × x^(−2) × y⁵
= 3y⁵ / x²
$

Answer: 3y⁵ / x²

Example 2: Negative and Fractional Indices Combined

Simplify and evaluate: 16^(−3/4) × 8^(2/3)

Solution:

$16^(−3/4) × 8^(2/3)
$

Step 1 — Handle 16^(−3/4):

$16^(−3/4) = 1 / 16^(3/4)
          = 1 / (16^(1/4))³       ← Root first
          = 1 / 2³                ← ⁴√16 = 2
          = 1/8
$

Step 2 — Handle 8^(2/3):

$8^(2/3) = (8^(1/3))²             ← Root first
        = 2²                     ← ∛8 = 2
        = 4
$

Step 3 — Multiply:

$(1/8) × 4 = 4/8 = 1/2
$

Answer: 1/2

Example 3: Solving an Exponential Equation

Solve for x: 5^(2x−1) × 25^(x) = 125

Solution:

$5^(2x−1) × 25ˣ = 125
5^(2x−1) × (5²)ˣ = 5³           ← 25 = 5², 125 = 5³
5^(2x−1) × 5^(2x) = 5³           ← Law 3: (5²)ˣ = 5^(2x)
5^(2x−1+2x) = 5³                  ← Law 1
5^(4x−1) = 5³
∴ 4x − 1 = 3
   4x = 4
   x = 1
$

Check: 5^(2(1)−1) × 25¹ = 5¹ × 25 = 5 × 25 = 125 ✓

Answer: x = 1

Example 4: Algebraic Expression with Fractional Indices

Simplify: (x^(1/2)y^(−1/3))⁶ × (x²y)^(1/3)

Solution:

$(x^(1/2) × y^(−1/3))⁶ × (x²y)^(1/3)
= x^((1/2)×6) × y^((−1/3)×6) × (x²)^(1/3) × y^(1/3)     ← Law 3 and Law 4
= x³ × y^(−2) × x^(2/3) × y^(1/3)                         ← Simplify each exponent
= x^(3 + 2/3) × y^(−2 + 1/3)                              ← Group same bases
= x^(11/3) × y^(−5/3)                                      ← 3 + 2/3 = 9/3 + 2/3 = 11/3; −2 + 1/3 = −6/3 + 1/3 = −5/3
= x^(11/3) / y^(5/3)                                       ← Convert negative index
$

Answer: x^(11/3) / y^(5/3)

Key Terms

• Law 2 — Quotient Of Powers: Aᵐ ÷ Aⁿ = A^(M−N) (A ≠ 0)
• Law 5 — Power Of A Quotient: (A / B)ⁿ = Aⁿ / Bⁿ (B ≠ 0)
• index
• Answer: B) 1/3
• Misplacing the negative sign.
• Example 4: Algebraic Expression With Fractional Indices
• 0⁰ is undefined
• Review — What Is An Index (Exponent)?
• What if m < n?
• Correct: A) x⁹
• Correct: B) 2
• Answer: A) k⁵

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Simplify: a⁵ × a³ ÷ a⁶

Problem 2: Simplify: (2p³q²)⁴ × (p²q⁻¹)³

Problem 3: Simplify to a single number: 27^(2/3) × 9^(−1/2)

Problem 4: Simplify and express with positive indices: (x⁻²y³)⁻¹ × (x⁴y⁻¹)²

Problem 5: Solve for x: 2ˣ × 4^(x+1) = 8²

Problem 6: Simplify: (x^(2/3)y²)³ ÷ (x⁻¹y^(1/2))⁴

Problem 7: Evaluate: (81^(−3/4) × 16^(1/2)) / 27^(−1/3)

Summary

1. Five fundamental laws: Product (add indices), Quotient (subtract indices), Power of power (multiply indices), Power of product and quotient (distribute index).
2. Zero index: a⁰ = 1 for any a ≠ 0, derived directly from the quotient law.
3. Negative index: a^(−n) = 1/aⁿ — it means "take the reciprocal." Always express final answers with positive indices.
4. Fractional index: a^(m/n) = (ⁿ√a)ᵐ — the denominator gives the root, the numerator gives the power. Take the root first for easier arithmetic.
5. Solving exponential equations: Express both sides with the same base, then equate the exponents. This is foundational for logarithms later.
6. Powers distribute over multiplication and division ONLY — (a+b)ⁿ ≠ aⁿ + bⁿ. This distinction is vital for all of algebra.

Pitfalls

• Adding indices instead of multiplying for power-of-power. (x³)² = x⁶, NOT x⁵.
• Multiplying indices instead of adding for product. x³ × x² = x⁵, NOT x⁶.
• Applying the product/quotient laws to different bases. 2³ × 3⁴ cannot be simplified with index laws — the bases must match.
• Misplacing the negative sign. a^(−n) = 1/aⁿ, NOT −aⁿ. 2^(−3) = 1/8, not −8.
• Forgetting that (a+b)ⁿ is NOT aⁿ + bⁿ. This error persists through years of algebra. (x + y)² = x² + 2xy + y², not x² + y².

Quiz

Answer each question, then read the explanation for your choice.

Q1: Simplify: x⁵ × x⁷ ÷ x³

A) x^(5+7−3) = x⁹ B) x^(5×7÷3) = x^(35/3) C) x^(5+7+3) = x¹⁵ D) x^(5×7−3) = x³²

Q2: What is (3x²y³)⁴?

A) 81x⁶y⁷ B) 12x⁶y⁷ C) 81x⁸y¹² D) 12x⁸y¹²

Q3: Simplify to a single number: (2⁶ × 2³) / (2⁴)²

A) 2⁵ B) 2 C) 2⁸ D) 2³

Q4: Express 5^(−2) as a positive fraction.

A) −25 B) 1/25 C) −1/25 D) 25

Q5: Evaluate 64^(2/3).

A) 16 B) 128/3 C) 8 D) 4

Q6: Simplify: (a³b⁻²)² × (a⁻¹b⁴)

A) a⁵b⁰ B) a⁵ C) a⁷b⁻⁸ D) a⁵ / b⁸

Q7: Solve for x: 3^(2x) = 27^(x−1)

A) x = 3 B) x = 2 C) x = −3 D) x = 1

Q8: Which is equivalent to x^(−1/2)?

A) 1/√x B) −√x C) √(−x) D) −1/√x

Next Steps

Move on to 00-12 — Scientific Notation and Standard Form to learn how to express very large and very small numbers compactly using powers of 10, and how to perform calculations in scientific notation.

Q5: Simplify: (x³y⁻²)⁻²

A) x⁻⁶y⁴ B) x⁶y⁻⁴ C) x⁻⁶ / y⁴ D) Both A and C

Answer: D) Both A and C (x³y⁻²)⁻² = x⁻⁶y⁴ = y⁴ / x⁶. Wait — x⁻⁶y⁴ is equivalent to y⁴/x⁶, not x⁻⁶/y⁴ (that would be 1/(x⁶y⁴)). Let me recheck: (x³y⁻²)⁻² = x^(3×(−2)) × y^((−2)×(−2)) = x⁻⁶ × y⁴ = y⁴ / x⁶. So A says x⁻⁶y⁴ which IS y⁴/x⁶ — yes that's correct. C says x⁻⁶ / y⁴ = 1/(x⁶y⁴) — NOT the same. So only A is correct. The answer should be A only. But D says "Both A and C" which is wrong. Actually, I'd realize both A and C being listed as "both" is confusing. Let me just say the answer is A. Or I can fix:

Actually, just to keep it clean: Answer: A) x⁻⁶y⁴ (= y⁴/x⁶)