Roots and Integer Indices
Index Laws — Integer Indices
When the same base appears repeatedly in multiplication or division, index laws allow us to simplify without evaluating every power. These laws are exact — they produce equivalent expressions, not approximations — and they extend through algebra, standard form, and beyond.
There are five core laws to know. Each applies only when the base is the same in all terms being combined. This condition is critical: the laws cannot be applied across different bases.
The Five Laws
| Law | General Form | Numerical Example |
|---|---|---|
| Multiplication | \( a^m \times a^n = a^{m+n} \) | \( 2^3 \times 2^4 = 2^7 = 128 \) |
| Division | \( a^m \div a^n = a^{m-n} \) | \( 5^6 \div 5^2 = 5^4 = 625 \) |
| Power of a Power | \( (a^m)^n = a^{mn} \) | \( (3^2)^3 = 3^6 = 729 \) |
| Zero Index | \( a^0 = 1 quad (a \neq 0) \) | \( 7^0 = 1,quad (-4)^0 = 1 \) |
| Negative Index | \( a^{-n} = \dfrac{1}{a^n} \) | \( 2^{-3} = \dfrac{1}{8},quad 10^{-2} = 0.01 \) |
Why Does the Multiplication Law Work?
Consider \( 2^3 \times 2^2 \). Writing out the repeated multiplication shows exactly why the powers add:
Why is \( a^0 = 1 \)?
Applying the division law with equal indices: \( a^n \div a^n = a^{n-n} = a^0 \). But any non-zero value divided by itself equals 1. Therefore \( a^0 = 1 \) for all \( a \neq 0 \). The case \( 0^0 \) is undefined.
Why is \( a^{-n} = \frac{1}{a^n} \)?
Applying the division law: \( a^0 \div a^n = a^{0-n} = a^{-n} \). Since \( a^0 = 1 \), this gives \( 1 \div a^n = \frac{1}{a^n} \). Negative indices represent reciprocals — they do not make the result negative.
Worked Examples
Simplify \( x^5 \times x^3 \) and \( 3a^4 \times 2a^2 \)
\[ x^5 \times x^3 = x^{5+3} = x^8 \]With a coefficient: multiply the number parts separately, then apply the index law to the letter parts.
\[ 3a^4 \times 2a^2 = (3 \times 2) \times a^{4+2} = 6a^6 \]Simplify \( \dfrac{y^9}{y^4} \) and \( \dfrac{12p^7}{4p^3} \)
\[ \frac{y^9}{y^4} = y^{9-4} = y^5 \] \[ \frac{12p^7}{4p^3} = \frac{12}{4} \times p^{7-3} = 3p^4 \]Simplify \( (a^3)^4 \) and \( (2x^2)^3 \)
\[ (a^3)^4 = a^{3 \times 4} = a^{12} \]When a bracket contains a coefficient, the power applies to everything inside:
\[ (2x^2)^3 = 2^3 \times (x^2)^3 = 8x^6 \]Evaluate \( 3^0 \), \( 5^{-2} \), and \( \left(\frac{1}{2}\right)^{-3} \)
\[ 3^0 = 1 \] \[ 5^{-2} = \frac{1}{5^2} = \frac{1}{25} \]For a fraction raised to a negative power, flip the fraction and change the sign of the index:
\[ \left(\frac{1}{2}\right)^{-3} = \left(\frac{2}{1}\right)^3 = 2^3 = 8 \]Key Takeaways
- Index laws only apply when bases are identical — never combine powers across different bases.
- Multiplication law: same base, add the indices. Division law: same base, subtract the indices.
- Power of a power: multiply the indices. A coefficient inside a bracket is also raised to that power.
- \( a^0 = 1 \) for all \( a \neq 0 \) — the zero index never makes a result zero.
- \( a^{-n} = \frac{1}{a^n} \) — a negative index means "reciprocal", not a negative result.