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YEAR 10 · EXPONENTIAL RELATIONSHIPS · 15 MIN

Integer and fractional index laws

Use index laws to simplify products, quotients and powers, including negative and fractional indices.

In this lesson

Before you start
  • Year 9 index laws
01 / LEARN

Understand the idea.

Identify the operation

First identify the operation between the powers. With a common base, a product adds indices and a quotient subtracts them: aman=am+n and aman=am−n, with a≠0 for division.

A power of a power multiplies indices: (am)n=am⁢n. Compare (a3)5=a15 with a3a5=a8: the brackets change which operation you perform. Addition of powers has no matching shortcut.

Zero, negative and fractional powers

For a≠0, a0=1 and a−n=1an. A negative index takes a reciprocal; it does not attach a minus sign to the answer. For instance, 2−3=18.

For a>0 and positive integer n, amn=(an)m. The denominator names the root and the numerator names the power. Using positive bases here keeps these index laws within the real numbers.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Combine two laws

Simplify x7x−2x3, where x≠0.

Worked solution

  1. 1

    Combine the numerator powers.

    x7x−2=x7+(−2)=x5

    Multiplication of equal bases adds their indices.

  2. 2

    Divide the remaining powers.

    x5x3=x5−3

    The nonzero-base condition permits subtraction of indices in a quotient.

  3. 3

    Simplify the exponent.

    x7x−2x3=x2(x≠0)

    The original restriction remains even though the final expression has no denominator.

Answer

x2

EXAMPLE 02

Evaluate a fractional power

Evaluate 163⁢/4.

Worked solution

  1. 1

    Read the fractional index.

    1634=(164)3

    The denominator gives a fourth root and the numerator gives a cube.

  2. 2

    Evaluate the root.

    164=2

    The principal fourth root is positive and 24=16.

  3. 3

    Cube the root.

    1634=23=8

    Taking the root first keeps the arithmetic small.

Answer

8

03 / YOUR TURN

Try it yourself.

Check your understanding

Simplify (b2)3b−1 for b≠0.

Show answer & reasoning

b5

First (b2)3=b6, then b6b−1=b6−1=b5. The condition b≠0 is needed for the negative power.

Keep practising

6 worksheet questions · 13 marks

Read the questions online
  1. Simplify x4x6.
    2 marks
  2. Simplify a8a3, a≠0.
    2 marks
  3. Simplify (m3)4.
    2 marks
  4. Evaluate 5−2.
    2 marks
  5. Evaluate 8134.
    2 marks
  6. Simplify p−2p5p4 for p≠0, using positive indices.
    3 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Simplify x4x6.
    2 marks

    Add indices for the common-base product: 4+6=10.

    x10
  2. Simplify a8a3, a≠0.
    2 marks

    8−3=5.

    a5
  3. Simplify (m3)4.
    2 marks

    Power of a power: 3×4=12.

    m12
  4. Evaluate 5−2.
    2 marks

    5−2=152=125.

    125
  5. Evaluate 8134.
    2 marks

    8114=3, so 8134=33=27.

    27
  6. Simplify p−2p5p4 for p≠0, using positive indices.
    3 marks

    −2+5−4=−1, so p−1=1p.

    1p
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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Integer and fractional index laws; connecting fractional powers with roots; simplifying surds; operations with surds; rationalising denominators; exact and approximate values; exponential tables and graphs; growth/decay factors; comparing linear/quadratic/exponential patterns; exponential equations; logarithms as inverses; logarithm laws; interpreting logarithmic scales; graph transformations and parameter experiments; choosing and evaluating models.

Official curriculum source ↗

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