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Module 7 (M7) – Number and algebra - Indices

Part of MathsM7: Number

Indices

It is helpful to read the guides from Module 6 (M6) on indices before starting this guide.

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Rules of indices when working with the same base

RULEALGEBRAIC RULENUMERICAL EXAMPLE
1. When multiplying, ADD the indices\(a^3 \times a^2 = a^5\)\(2^3 \times 2^2 = 2^5 = 32\)
2. When dividing, SUBTRACT the indices\(a^5 \div a^2 = a^3\)\(3^5 \div 3^2 = 3^3 = 27\)
3. When the power is raised to another power, MULTIPLY the indices.\((a^3)^2 = a^6\)\((3^3)^2 = 3^6 = 729\)
4. Anything to the power of 0 is 1\(a^0  = 1\)\(43^0  = 1\)

We can add a further rule for negative indices.

RULEALGEBRAIC RULENUMERICAL EXAMPLE
5. Negative indices indicate a fraction\(a^{-n}= \frac{1}{a^{n}}\)\(3^{-2}=\ \frac{1}{3^{2}} = \frac{1}{9}\)

Often, more than one rule has to be used when simplifying expressions with indices.

Example

Simplify \(p^3 \div p^{6}\)

Solution

  • Using Rule 2:

    \(p^{3}\div p^{6} = p^{3–6}\)

    \(p^{–3}\)

  • Using Rule 5:

    \(p^{–3} = \frac{1}{p^{3}}\)

Answer

\(\frac{1}{p^{3}}\)

Question

Simplify \((a^{3})^{2}\div a^{7}\).

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Solving equations using rules for indices

Example

Find the value of \(n\) if \((\frac{1}{3})^{n} = 81\)

Solution

The base is 3, so rearrange both sides as powers of 3.

Left hand side of the equation
  • Using Rule 3

\((\frac{1}{3})^{n} = (3^{–1})^{n}\)

\( = 3^{–n}\)

Right hand side of the equation
  • Write the equation as powers of 3

\(81 = 3^{4}\)

  • Rewrite the equation as powers of 3

\( = 3^{–n} = 3^{4}\)

  • Equate indices

\( = –n = 4\)

\( = n = –4\)

Answer

\(n = –4\)

Question

Find the value of \(x\) if \(2^{x–1}=\frac{1}{16}\).

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