An index counts how many times a number is multiplied by itself. Three laws handle multiplying, dividing and raising powers; zero and negative indices follow from them, and (Extended) fractional indices are roots.
With the same base: \(a^{m} \times a^{n} = a^{m+n}\), \(a^{m} \div a^{n} = a^{m-n}\) and \((a^{m})^{n} = a^{mn}\). Also \(a^{0} = 1\), \(a^{-n} = \dfrac{1}{a^{n}}\) and (Extended) \(a^{\frac{1}{n}} = \sqrt[n]{a}\).
Fig. 1: why the indices add when you multiply: 2³ × 2² is five 2s multiplied together, so it is 2⁵ = 32
Section 01
The three laws.
In the lab, choose a base \(a\) and the powers \(m\) and \(n\), and watch the factors line up: multiplying joins the two strings of factors, dividing cancels them in pairs. Predict the index before you look.
Law
Example
\(a^{m} \times a^{n} = a^{m+n}\)
\(3^{4} \times 3^{2} = 3^{6}\)
\(a^{m} \div a^{n} = a^{m-n}\)
\(5^{7} \div 5^{3} = 5^{4}\)
\((a^{m})^{n} = a^{mn}\)
\((2^{3})^{2} = 2^{6}\)
Worked example Using the laws
Find the value of (a) \(2^{-3} \times 2^{4}\) (b) \((2^{3})^{2}\) (c) \(2^{3} \div 2^{4}\).
(a) Same base, multiplying: add the indices, \(-3 + 4 = 1\), so \(2^{1} = 2\).
(b) Power of a power: multiply the indices, \(3 \times 2 = 6\), so \(2^{6} = 64\).
(c) Dividing: subtract the indices, \(3 - 4 = -1\), so \(2^{-1} = \dfrac{1}{2}\).
Answer: (a) 2 (b) 64 (c) \(\dfrac{1}{2}\)
Worked example Leaving the answer as a power
Simplify \(\dfrac{7^{5} \times 7^{3}}{7^{6}}\), leaving your answer in index form.
Top: \(7^{5} \times 7^{3} = 7^{8}\).
Divide: \(7^{8} \div 7^{6} = 7^{2}\).
Answer: \(7^{2}\) (which is 49)
Examiner trap
The laws only work with the SAME base, and the base never changes: \(3^{4} \times 3^{2} = 3^{6}\), not \(9^{6}\). And \(2^{3} \times 5^{2}\) cannot be combined at all.
Section 02
Zero and negative indices.
Dividing \(5^{3}\) by itself gives 1, but the law gives \(5^{3-3} = 5^{0}\), so \(5^{0} = 1\). Dividing \(2^{3}\) by \(2^{5}\) gives \(\dfrac{1}{4}\), and the law gives \(2^{-2}\), so a negative index means 'one over'.
\(a^{0} = 1\) for any non-zero \(a\): \(7^{0} = 1\), \((-3)^{0} = 1\).
\(81 = 3^{4}\), so \(\dfrac{1}{81} = \dfrac{1}{3^{4}} = 3^{-4}\).
Compare indices: \(n = -4\).
Answer: \(n = -4\)
Examiner trap
A negative index does NOT make the number negative: \(7^{-2} = \dfrac{1}{49}\), not \(-49\) and not \(-14\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Simplify \(5^{3} \times 5^{4}\).
Quick check+10 XP
Find the value of \(7^{0}\).
Quick check+10 XP
Find the value of \(2^{-3}\).
Quick check+10 XP
Simplify \((3^{2})^{4}\).
Section 03
Fractional indices.
Extended. Since \(\left(9^{\frac{1}{2}}\right)^{2} = 9^{1} = 9\), the number \(9^{\frac{1}{2}}\) must be \(\sqrt{9} = 3\). The bottom of a fractional index is a root, and the top is a power.