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YEAR 11 · EXPONENTIALS AND LOGARITHMS · 15 MIN

Integer and fractional index laws

Understand roots and reciprocals, then apply index laws with appropriate restrictions.

In this lesson

Before you start
  • Integer powers and equivalent fractions
01 / LEARN

Understand the idea.

Read the exponent first

An exponent tells you how a base is used. A positive whole-number exponent repeats multiplication; a negative exponent takes the reciprocal of the corresponding positive power. It does not put a minus sign in front of the value.

For a positive base, the denominator of a fractional exponent specifies a root and the numerator specifies a power. When the root is exact, taking it first often avoids large intermediate numbers.

am⁢/⁢n=(an)m(a>0)

Combine structure before calculating

Check the structure before applying an index law: multiplying equal bases adds exponents, dividing them subtracts exponents, and taking a power of a power multiplies exponents. Brackets show which complete expression the exponent acts on.

Sign restrictions are part of the method. The principal square root is nonnegative, so x2=∣⁢x⁢∣. The condition x>0 in these fractional-power examples allows roots and powers to be simplified without sign ambiguity.

aman=am+n(am)n=am⁢n
02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A fractional power

Evaluate 813⁢/4.

Worked solution

  1. 1

    Interpret the fractional exponent.

    8134=(814)3

    The denominator gives the root and the numerator gives the subsequent power.

  2. 2

    Take the exact fourth root.

    814=3

    The positive fourth root is 3 because 34=81.

  3. 3

    Cube the result.

    8134=33=27

    The numerator of the fractional exponent requires the third power.

Answer

27

EXAMPLE 02

A powered product

Simplify (16x8)3⁢/4 for x>0.

Worked solution

  1. 1

    Apply the exponent to both positive factors.

    (16x8)34=1634(x8)34

    The condition x>0 makes the fractional-power product law valid here.

  2. 2

    Evaluate the number and combine the variable powers.

    1634=23=8,(x8)34=x8⁢(34)=x6

    A power of a power multiplies its exponents.

  3. 3

    Combine the simplified factors.

    (16x8)34=8x6

    The numerical coefficient and variable factor are now both in simplest power form.

Answer

8x6

03 / YOUR TURN

Try it yourself.

Check your understanding

Evaluate 27−2⁢/3.

Show answer & reasoning

1⁢/9

27−23=1(273)2=132=19. The negative exponent requires the reciprocal after evaluating the positive power.

Keep practising

6 worksheet questions · 14 marks

Read the questions online
  1. Evaluate 2−3.
    2 marks
  2. Evaluate 3225.
    2 marks
  3. Evaluate 16−34.
    2 marks
  4. Simplify a4a3a2 for a≠0.
    2 marks
  5. Simplify (16x4)34 for x>0.
    3 marks
  6. Check the claim (9+16)12=912+1612 and explain the error.
    3 marks
Read the worked solutions

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

  1. Evaluate 2−3.
    2 marks

    2−3=123=18.

    18
  2. Evaluate 3225.
    2 marks

    325=2, then 22=4.

    4
  3. Evaluate 16−34.
    2 marks

    164=2; cube to get 8 and invert.

    18
  4. Simplify a4a3a2 for a≠0.
    2 marks

    The exponent is 4+3−2=5.

    a5
  5. Simplify (16x4)34 for x>0.
    3 marks

    1634=8; 4⁢(34)=3.

    8x3
  6. Check the claim (9+16)12=912+1612 and explain the error.
    3 marks

    25=5 while 9+16=3+4=7.

    A product power law does not apply to addition.

    False: the sides are 5 and 7.
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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

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