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What are the laws of indices, and how do we use them to simplify expressions and solve index equations?

State and apply the laws of indices, including zero, negative and fractional indices, to simplify expressions and solve simple index equations

A focused answer to the N(A)-Level Additional Mathematics outcome on indices. The laws of indices including zero, negative and fractional powers, how to simplify expressions, and how to solve simple equations by matching bases.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to know the laws of indices and use them fluently to simplify expressions involving powers, including zero, negative and fractional indices. You also need to solve simple index equations such as 2x=322^x = 32 by writing both sides with the same base. Indices are the language of growth and decay and the foundation for logarithms, so this is one of the most reused skills in the whole syllabus.

The answer

The core laws

For the same base aa (with a0a \ne 0):

am×an=am+n,aman=amn,(am)n=amna^m \times a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{mn}

When you multiply like bases you add indices; when you divide you subtract; a power of a power multiplies. Powers also distribute over a product: (ab)n=anbn(ab)^n = a^n b^n.

Zero and negative indices

a0=1(a0),an=1ana^0 = 1 \quad (a \ne 0), \qquad a^{-n} = \frac{1}{a^n}

A zero index gives 11; a negative index means "one over". For example 50=15^0 = 1 and 23=123=182^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}.

Fractional indices

A fractional index is a root combined with a power:

a1/n=an,am/n=(an)m=amna^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^m}

So 91/2=9=39^{1/2} = \sqrt{9} = 3 and 82/3=(83)2=22=48^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4. Taking the root first usually keeps the numbers small.

Rewriting roots and reciprocals as indices

A great deal of algebra becomes easier once roots and reciprocals are written as indices, because the three core laws can then be applied directly. The standard conversions are:

x=x1/2,1x=x1,1x=x1/2,1xn=xn\sqrt{x} = x^{1/2}, \qquad \frac{1}{x} = x^{-1}, \qquad \frac{1}{\sqrt{x}} = x^{-1/2}, \qquad \frac{1}{x^n} = x^{-n}

For example, x2x=x2x1/2=x21/2=x3/2\dfrac{x^2}{\sqrt{x}} = \dfrac{x^2}{x^{1/2}} = x^{2 - 1/2} = x^{3/2}. Converting first turns an awkward-looking fraction into a single power that obeys the subtraction law cleanly.

Solving simple index equations

If you can write both sides of an equation as powers of the same base, then the indices must be equal:

af(x)=ag(x)f(x)=g(x)a^{f(x)} = a^{g(x)} \quad\Rightarrow\quad f(x) = g(x)

This works because the exponential function is one-to-one. When the bases cannot be matched, logarithms are used instead (covered in the exponential and logarithmic strand).

Examples in context

Example 1. Compound growth. A sum of money growing at a fixed rate is modelled by A=P(1.05)nA = P(1.05)^n after nn years. Manipulating that power, and later solving for nn, depends entirely on the index laws, which is why indices lead directly into the logarithm topic.

Example 2. Simplifying before differentiating. Calculus questions often hide a power, for example 1x=x1/2\dfrac{1}{\sqrt{x}} = x^{-1/2}. Rewriting roots and reciprocals as indices first is what makes the power rule for differentiation applicable, so this skill resurfaces in the calculus strand.

Try this

Q1. Simplify (x3)2×x(x^3)^2 \times x. [2 marks]

  • Cue. (x3)2=x6(x^3)^2 = x^6, then x6×x=x7x^6 \times x = x^7.

Q2. Evaluate 272/327^{2/3}. [2 marks]

  • Cue. (273)2=32=9\left(\sqrt[3]{27}\right)^2 = 3^2 = 9.

Q3. Solve 3x=193^{x} = \dfrac{1}{9}. [2 marks]

  • Cue. 19=32\dfrac{1}{9} = 3^{-2}, so x=2x = -2.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksSimplify x5×x2x3\dfrac{x^5 \times x^2}{x^3}, giving your answer as a single power of xx.
Show worked answer →

Multiply by adding indices on top: x5×x2=x5+2=x7x^5 \times x^2 = x^{5+2} = x^7.

Divide by subtracting the index: x7x3=x73=x4\dfrac{x^7}{x^3} = x^{7-3} = x^4.

What markers reward: adding indices when multiplying like bases, subtracting when dividing, and giving a single simplified power x4x^4.

Original3 marksSolve 2x=322^x = 32.
Show worked answer →

Write 3232 as a power of the same base: 32=2532 = 2^5.

So 2x=252^x = 2^5, and since the bases are equal the indices are equal: x=5x = 5.

What markers reward: expressing both sides with the same base, equating the indices once the bases match, and stating x=5x = 5.

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