1. The rules of indices
Working with the same base
- The index rules apply to algebra exactly as they do to numbers, and the multiplying and dividing rules need the terms to share a base
| Rule | What it does | Example |
|---|---|---|
| aᵐ × aⁿ = aᵐ⁺ⁿ | Multiplying adds the indices | x⁵ × x³ = x⁸ |
| aᵐ ÷ aⁿ = aᵐ⁻ⁿ | Dividing subtracts the indices | y⁹ ÷ y⁴ = y⁵ |
| (aᵐ)ⁿ = aᵐⁿ | A power of a power multiplies them | (m³)⁴ = m¹² |
| (ab)ⁿ = aⁿbⁿ | The power reaches every factor | (2a³)⁴ = 16a¹² |
| a⁰ = 1 | Any non-zero base to the power zero | 7x⁰ = 7 |
- A coefficient is raised to the power along with the letter, which is what turns 2 into 16 in the example above
- Terms with different bases cannot be combined by these rules, so x³ × y² stays as it is
Worked example
Multiplying terms with numbers in front
Simplify 6m⁴n × 3m²n³.
Solution:
- Deal with the numbers and with each letter separately
- Multiply the numbers: 6 × 3 = 18. They are multiplied, not added
- Add the powers of m: m⁴ × m² = m⁶
- n on its own is n¹, so add the powers of n: n¹ × n³ = n⁴
- The answer is 18m⁶n⁴
Worked example
Applying the index rules
Simplify x⁵ × x⁻² ÷ x⁴, and work out the value of 16^(3/4).
Solution:
- Multiplying powers of the same base adds the indices: 5 + (−2) = 3, giving x³
- Dividing subtracts them: 3 − 4 = −1, giving x⁻¹
- A negative index means a reciprocal, so the answer is 1/x
- For 16^(3/4), the denominator of the fraction is the root and the numerator is the power
- Take the fourth root first, because it keeps the numbers small: ⁴√16 = 2
- Then raise to the power 3: 2³ = 8
- Doing it the other way round means cubing 16 to 4096 first, which is the same answer by a much harder route
2. Negative and fractional indices
Negative indices
- A negative index means the reciprocal of the positive power
- p⁻³ = 1/p³
- The sign of the index says nothing about the sign of the value
Fractional indices
- A unit fraction index is a root, with the denominator giving which root
- a^(1/2) is the square root and a^(1/3) is the cube root
- A general fraction m/n combines both: take the nth root, then raise to the power m
- 8^(2/3) is the cube root of 8, which is 2, then squared, giving 4
- Rooting before powering keeps the numbers small, which matters without a calculator
- A negative fractional index combines the two ideas, giving one over the root
Worked example
Simplifying a power of a product
Simplify (16x⁴)^(1/2).
Solution:
- The power applies to both the number and the letter
- The square root of 16 is 4
- For x⁴, halve the index: x⁴ raised to the power 1/2 gives x²
- The simplified form is 4x²
Worked example
A negative fractional power of a bracket
Simplify (81x⁸)^(−3/4).
Solution:
- The power applies to the 81 and to the x⁸ separately
- For the number, take the fourth root first: ⁴√81 = 3
- Then cube it: 3³ = 27, and the negative sign makes it the reciprocal, 1/27
- For the letter, multiply the indices: 8 × (−3/4) = −6, giving x⁻⁶ = 1/x⁶
- Putting the two parts together gives 1/(27x⁶), which can also be written (1/27)x⁻⁶
3. Solving equations with indices
Matching the bases
- Where an unknown appears in the index, rewrite both sides using the same base, then equate the indices
- This works because a single base raised to two powers is equal only when the powers match
- Recognising powers of 2, 3, 5 and 7 is what makes the rewriting possible
Worked example
Solving an index equation
Solve 2^(x + 3) = 8ˣ.
Solution:
- Write 8 as a power of 2: 8 = 2³
- The right-hand side becomes (2³)ˣ, which is 2^(3x)
- Both sides now share base 2, so the indices are equal: x + 3 = 3x
- Rearranging gives 3 = 2x, so x = 1.5
Worked example
Changing both sides to a shared base
Solve (1/4)ˣ = 8^(x + 1).
Solution:
- 1/4 and 8 are both powers of 2, so write each side with base 2
- One over a power is a negative power: 1/4 = 1/2² = 2⁻², so (1/4)ˣ = (2⁻²)ˣ = 2^(−2x)
- 8 = 2³, so 8^(x + 1) = (2³)^(x + 1) = 2^(3(x + 1))
- The 3 multiplies the whole index, giving 3x + 3, not 3x + 1
- The bases now match, so the indices are equal: −2x = 3x + 3
- Then −5x = 3, so x = −3/5
- Where the base carrying the unknown is the larger power, the answer is often a fraction
- 27ˣ = 3 becomes 3^(3x) = 3¹, so 3x = 1 and x = 1/3
- Logarithms are not required by this specification, so every such equation can be solved by matching bases, unless a question asks you to solve it from a graph (2ˣ = 5, say, has no shared base)