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How do the laws of indices let us simplify expressions with positive, negative, zero and fractional powers?

Apply the laws of indices to simplify expressions involving positive, negative, zero and fractional powers and to solve simple exponential equations

A focused answer to the O-Level A-Maths outcome on indices. The laws of exponents, the meaning of zero, negative and fractional powers, and using them to simplify and to solve simple index equations.

Generated by Claude Opus 4.88 min answer

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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 handle powers fluently: to multiply, divide and raise powers using the index laws, and to read negative, zero and fractional indices correctly. Indices underpin surds (a fractional index is a root), logarithms (the inverse operation), and the whole of differentiation and integration of xnx^n, so this is foundational.

The answer

The multiplication, division and power laws

For the same base aa:

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

Add indices when multiplying, subtract when dividing, and multiply indices when raising a power to a power.

Powers of products and quotients

A power distributes over a product or a quotient:

(ab)n=anbn,(ab)n=anbn.(ab)^n = a^n b^n, \qquad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}.

So (2x2)3=23x6=8x6(2x^2)^3 = 2^3 x^6 = 8x^6; the index applies to every factor, including the number.

Zero, negative and fractional indices

These three definitions extend the laws to all rational powers:

a0=1 (a≠0),a−n=1an,a1/n=an,am/n=(an)m.a^0 = 1\ (a \neq 0), \qquad a^{-n} = \frac{1}{a^n}, \qquad a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^m.

A negative index means a reciprocal; a fractional index means a root. So 82/3=(83)2=22=48^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4.

Solving simple exponential equations

If you can write both sides of an equation to the same base, then equal powers force equal indices: ax=ay⇒x=ya^x = a^y \Rightarrow x = y. This turns an index equation into an ordinary algebraic one.

Common bases to recognise

It helps to know the small powers of the usual bases: 4=224 = 2^2, 8=238 = 2^3, 16=2416 = 2^4, 9=329 = 3^2, 27=3327 = 3^3, 25=5225 = 5^2. Spotting these lets you rewrite both sides of an equation such as 8x=48^x = 4 to a common base 22 and equate indices at once.

The link to surds and logarithms

A fractional index is exactly a surd (x1/2=xx^{1/2} = \sqrt{x}), so the index laws and surd rules are two views of the same thing. When the bases cannot be matched, the inverse operation, the logarithm, is needed, which is the natural next topic after indices.

Examples in context

Example 1. Compound growth. A sum invested at 5%5\% per year grows by a factor 1.05n1.05^n after nn years. The index laws let you combine or compare such factors, and the same exponential form reappears in the logarithm and exponential-function topics.

Example 2. Preparing to differentiate. Before differentiating y=1x3+xy = \dfrac{1}{x^3} + \sqrt{x}, you rewrite it as y=x−3+x1/2y = x^{-3} + x^{1/2} using negative and fractional indices, so that the power rule ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1} can be applied directly.

Try this

Q1. Simplify a5×a−2÷a0a^5 \times a^{-2} \div a^{0}. [2 marks]

  • Cue. a5+(−2)−0=a3a^{5 + (-2) - 0} = a^{3}.

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

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

Q3. Solve 4x=84^{x} = 8. [2 marks]

  • Cue. Base 22: 22x=232^{2x} = 2^3, so 2x=32x = 3 and x=32x = \dfrac{3}{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 (2x2)3×x−4x2\dfrac{(2x^2)^3 \times x^{-4}}{x^2}, giving your answer with a positive index.
Show worked answer →

Expand the bracket: (2x2)3=23x6=8x6(2x^2)^3 = 2^3 x^{6} = 8x^{6}.

Multiply and divide using the index laws: 8x6×x−4x2=8x6−4x2=8x2x2=8x0=8\dfrac{8x^{6} \times x^{-4}}{x^2} = \dfrac{8x^{6 - 4}}{x^2} = \dfrac{8x^{2}}{x^2} = 8x^{0} = 8.

Markers reward cubing both the 22 and the index, adding indices when multiplying, subtracting when dividing, and recognising x0=1x^0 = 1.

Original3 marksSolve 9x=27 x−19^{x} = 27^{\,x - 1}.
Show worked answer →

Write both sides to the base 33: 9=329 = 3^2 and 27=3327 = 3^3.

So 32x=33(x−1)3^{2x} = 3^{3(x - 1)}. Equate the indices: 2x=3(x−1)=3x−32x = 3(x - 1) = 3x - 3.

Then −x=−3-x = -3, so x=3x = 3.

Markers reward expressing both sides to a common base, equating indices once the bases match, and solving the linear equation.

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