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A&G-Indices II
Indices (Laws of Exponents)
Complete algebra guide with exam‑style examples
What are indices?
An index (or exponent) shows how many times a number (the base) is multiplied by itself.
an = a × a × … × a (n times)
where a is the base and n is the index/exponent.
Example: 25 = 2 × 2 × 2 × 2 × 2 = 32.
Quick reference – Laws of Indices
Multiplication: am × an = am+n
Division: am ÷ an = am-n
Power of power: (am)n = am×n
Negative index: a-n = 1 / an
Zero index: a0 = 1 (for a ≠ 0)
Fractional index: am/n = n√(am)
1. Multiplication law
am × an = am+n
When multiplying powers with the same base, add the indices.
Example 1: Simplify x3 × x5.
x3 × x5 = x3+5 = x8.
Example 2: Simplify 2y2 × 3y4.
Multiply coefficients: 2 × 3 = 6; add indices: y2+4 = y6.
Answer: 6y6.
2. Division law
am ÷ an = am-n (for a ≠ 0)
When dividing powers with the same base, subtract the indices.
Example: Simplify x7 / x3 .
x7 ÷ x3 = x7-3 = x4.
Remember: am ÷ an = am/an. Works for any base except zero.
3. Power of a power
(am)n = am × n
Raise a power to another power by multiplying the indices.
Example: Simplify (23)4.
(23)4 = 23×4 = 212.
Example with variable: Simplify (x5)2.
(x5)2 = x5×2 = x10.
4. Negative and zero indices
Zero index
a0 = 1 for any a ≠ 0
Example: 50 = 1; (-3)0 = 1.
Negative index
a-n = 1 / an (for a ≠ 0)
A negative index means "one over" the positive power.
Example 1: Write 2-3 as a fraction.
2-3 = 1 / 23 = 1/8.
Example 2: Simplify 1 / x-2.
1 / x-2 = x2 because x-2 = 1/x2, so its reciprocal is x2.
5. Fractional indices
Fractional indices represent roots and powers.
a1/n = n√a (the n‑th root)
am/n = n√(am) = (n√a)m
am/n = n√(am) = (n√a)m
Example 1: Evaluate 91/2.
91/2 = √9 = 3.
Example 2: Simplify 272/3.
272/3 = (∛27)2 = 32 = 9.
Example 3: Write 4√(x3) using a fractional index.
4√(x3) = x3/4.
6. Solving equations with indices
Often we need to find the value of the base or the exponent.
Type 1: Same base
If am = an then m = n (provided a ≠ 0,1,–1).
Example: Solve 2x = 32.
32 = 25, so 2x = 25 → x = 5.
Type 2: Change to common base
Example: Solve 9x = 27.
Write both sides with base 3: 9 = 32 so 9x = (32)x = 32x.
27 = 33.
Therefore 32x = 33 → 2x = 3 → x = 1.5.
Type 3: Using substitution (quadratic form)
Example: Solve 22x – 5·2x + 4 = 0.
Let u = 2x. Then 22x = (2x)2 = u2.
Equation becomes u2 – 5u + 4 = 0 → (u–1)(u–4)=0.
So u = 1 or u = 4.
2x = 1 → x = 0; 2x = 4 → 2x = 22 → x = 2.
Solutions: x = 0 or x = 2.
Summary of index laws
| Law | Example |
|---|---|
| am × an = am+n | x3 · x4 = x7 |
| am ÷ an = am-n | y8 / y2 = y6 |
| (am)n = amn | (23)4 = 212 |
| a0 = 1 | 170 = 1 |
| a-n = 1 / an | 5-2 = 1/25 |
| a1/n = n√a | 161/4 = 2 |
| am/n = n√(am) | 82/3 = 4 |
Key exam tips
- Always simplify coefficients separately from the variables.
- Be careful with negative indices – they mean reciprocal, not negative numbers.
- Fractional indices: the denominator is the root, the numerator is the power.
- When solving equations, try to express both sides with the same base.
- Check for extraneous solutions if you square both sides or use substitution.
- Remember that a0 = 1 only if a ≠ 0. 00 is undefined.
Practice tip: Work through past exam questions – indices appear in almost every algebra paper.
Indices: Simplification problems
Simplify each expression using the laws of indices. All variables represent non‑zero numbers.
1.
x5 × x3
x5+3 = x8
2.
(y4)2
y4×2 = y8
3.
2a3 × 3a5
2×3 × a3+5 = 6a8
4.
(2x2y3)4
24 × (x2)4 × (y3)4 = 16x8y12
5.
12x7y4 / 4x3y
(12/4) × x7-3 × y4-1 = 3x4y3
6.
(a2b-3)2 × a5b4
a4b-6 × a5b4 = a4+5 b-6+4 = a9b-2 = a9 / b2
7.
5x-2y3 / 15x4y-1
(5/15) × x-2-4 × y3-(-1) = 1/3 × x-6 × y4 = y4 / (3x6)
8.
(8x6)2/3
82/3 × (x6)2/3 = (∛8)2 × x6×(2/3) = 22 × x4 = 4x4
9.
2x1/2 y3 / 4x-1/2 y
(2/4) × x1/2 - (-1/2) × y3-1 = 1/2 × x1 × y2 = (x y2)/2
10.
(9a-4b2)-1/2 × (a3b-1)2
First term: (9a-4b2)-1/2 = 9-1/2 × a(-4)×(-1/2) × b2×(-1/2)
= 1/√9 × a2 × b-1 = 1/3 a2 b-1.
Second term: (a3b-1)2 = a6 b-2.
Multiply: (1/3 a2 b-1) × (a6 b-2) = 1/3 a2+6 b-1-2 = 1/3 a8 b-3 = a8 / (3b3).