Indices in IGCSE Maths
The six index laws, negative and fractional indices, and how to apply them to simplify expressions and solve equations.
0580 · E1.7 / E2.4What are indices?
An index (plural: indices) is the small number written above and to the right of a base number. It tells you how many times to multiply the base by itself.
\[a^n = \underbrace{a \times a \times a \times \cdots \times a}_{n \text{ times}}\]In the expression \(2^5\), the base is 2 and the index (or power/exponent) is 5. So \(2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32\).
The six index laws
| Law | In words | Rule | Example |
|---|---|---|---|
| Multiplying | Add the powers | \(a^m \times a^n = a^{m+n}\) | \(x^3 \times x^4 = x^7\) |
| Dividing | Subtract the powers | \(a^m \div a^n = a^{m-n}\) | \(x^8 \div x^3 = x^5\) |
| Power of a power | Multiply the powers | \((a^m)^n = a^{mn}\) | \((x^2)^5 = x^{10}\) |
| Power of a product | Apply the power to each factor | \((ab)^n = a^n b^n\) | \((3x)^2 = 9x^2\) |
| Power of a fraction | Apply the power to top and bottom | \(\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}\) | \(\left(\dfrac{x}{2}\right)^3 = \dfrac{x^3}{8}\) |
| Zero index | Anything to the power 0 is 1 | \(a^0 = 1\) | \(7^0 = 1\) |
The first three laws only work when the bases are the same. You can simplify \(x^3 \times x^4\) because both bases are \(x\). You cannot simplify \(x^3 \times y^4\) using index laws — the bases are different.
Simplify \(3x^4 \times 5x^3\).
Multiply the numbers, then add the indices:
\[3x^4 \times 5x^3 = 15x^{4+3} = 15x^7\]Simplify \(\dfrac{12x^5}{4x^2}\).
Divide the numbers, then subtract the indices:
\[\frac{12x^5}{4x^2} = 3x^{5-2} = 3x^3\]Simplify \((2x^3)^4\).
Raise the number to the power, and multiply the indices:
\[(2x^3)^4 = 2^4 \times x^{3 \times 4} = 16x^{12}\]The zero index
Anything raised to the power of zero equals 1:
\[a^0 = 1 \qquad \text{(provided } a \neq 0\text{)}\]Using the division law: \(\dfrac{a^n}{a^n} = a^{n-n} = a^0\). But any number divided by itself is 1. So \(a^0 = 1\).
\((3x)^0 = 1\) because the whole bracket is raised to the power 0. But \(3x^0 = 3 \times x^0 = 3 \times 1 = 3\) because only \(x\) has the power 0.
Negative indices
A negative index means "one over" (the reciprocal):
\[a^{-n} = \frac{1}{a^n}\]A negative index does not make the answer negative. It flips the base to the bottom of a fraction.
Write \(\dfrac{3}{x^2}\) using a negative index.
The \(x^2\) in the denominator becomes \(x^{-2}\) in the numerator.
A negative index on a fraction flips the fraction, then applies the positive index:
\[\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}\]For example, \(\left(\dfrac{2}{5}\right)^{-3} = \left(\dfrac{5}{2}\right)^{3} = \dfrac{125}{8}\). This principle also appears when finding perpendicular gradients.
Fractional indices
A fractional index means a root (see surds). The denominator tells you which root:
\[a^{1/n} = \sqrt[n]{a}\]When the numerator isn't 1, it means "root and power":
\[a^{m/n} = (\sqrt[n]{a})^m\]You can do the root first or the power first — the answer is the same. Doing the root first usually keeps the numbers smaller.
Evaluate \(8^{\frac{2}{3}}\).
So \(8^{\frac{2}{3}} = 4\).
Evaluate \(27^{-\frac{2}{3}}\).
So \(27^{-\frac{2}{3}} = \dfrac{1}{9}\).
When you see a negative fractional index like \(a^{-\frac{m}{n}}\), deal with the three parts in this order: negative → root → power. Flip it, root it, power it.
Combining the laws
Simplify \(\dfrac{6x^5 y^3}{2x^2 y}\).
\(6 \div 2 = 3\)
\(x^5 \div x^2 = x^{5-2} = x^3\)
\(y^3 \div y^1 = y^{3-1} = y^2\)
Simplify \(\left(16x^8\right)^{\frac{1}{2}}\).
\(16^{\frac{1}{2}} = \sqrt{16} = 4\)
\(x^{8 \times \frac{1}{2}} = x^4\)
Solve \(2^x = 32\).
\(32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5\)
If the bases are the same, the indices must be equal:
\[\begin{aligned}2^x &= 2^5 \\[6pt] x &= 5\end{aligned}\]Solve \(3^{2x+1} = 27\).
\(27 = 3 \times 3 \times 3 = 3^3\)
Exam-style questions
Simplify \(\dfrac{4x^3 y^5}{8x y^2}\).
Show solution
Evaluate \(\left(\dfrac{3}{4}\right)^{-2}\).
Show solution
Flip the fraction, then apply the positive power:
\[\left(\frac{3}{4}\right)^{-2} = \left(\frac{4}{3}\right)^{2} = \frac{16}{9}\]Evaluate \(125^{\frac{2}{3}}\).
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Root first: \(\sqrt[3]{125} = 5\). Then power: \(5^2 = 25\).
\[125^{\frac{2}{3}} = 25\]Simplify \(\left(\dfrac{27x^6}{8}\right)^{\frac{2}{3}}\).
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Apply the power to each part of the fraction:
\[\begin{aligned}\left(\frac{27x^6}{8}\right)^{\frac{2}{3}} &= \frac{27^{\frac{2}{3}} \times x^{6 \times \frac{2}{3}}}{8^{\frac{2}{3}}} \\[10pt] &= \frac{(\sqrt[3]{27})^2 \times x^4}{(\sqrt[3]{8})^2} \\[10pt] &= \frac{9x^4}{4}\end{aligned}\]Solve \(4^x = 8\).
Show solution
Write both sides as powers of 2:
\[\begin{aligned}(2^2)^x &= 2^3 \\[6pt] 2^{2x} &= 2^3 \\[6pt] 2x &= 3 \\[6pt] x &= \frac{3}{2}\end{aligned}\]Common mistakes
\(x^3 \times x^4 = x^7\), not \(x^{12}\). When multiplying with the same base, you add the indices. You only multiply indices for a power of a power: \((x^3)^4 = x^{12}\).
\(2^{-3} = \dfrac{1}{8}\), not \(-8\). A negative index means reciprocal, not a negative number. The answer is always positive (if the base is positive).
In \((2x^3)^4\), the 2 also gets raised to the power: \(2^4 \times x^{12} = 16x^{12}\), not \(2x^{12}\). Everything inside the bracket is affected by the power.
You cannot simplify \(2^3 \times 3^4\) using index laws — the bases (2 and 3) are different. The laws only apply when the bases match. Similarly, \(x^3 \times y^2 \neq (xy)^5\).
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