Fractional Indices

a^{\dfrac{1}{2}}=\sqrt{a} (square root).

a^{\dfrac{1}{3}}=\sqrt[3] {a} (cube root)

a^{\dfrac{1}{n}}=\sqrt[n] {a} (nth root)

a^{\dfrac{m}{n}}=\left( a^{m}\right) ^{\dfrac{1}{n}}=\sqrt[n] {a^{m}}

Example

Simplify the following:

1. 8^{\dfrac{2}{3}}

2. 16^{\dfrac{1}{4}}

3. 125^{-\dfrac{2}{3}}

4. 64^{-\dfrac{4}{3}}

5. 36^{-\dfrac{1}{2}}

6. 81^{\dfrac{5}{4}}

7. \left( \dfrac{2}{3}\right) ^{-2}

8. -8^{\dfrac{2}{3}}

9. -27^{-\dfrac{4}{3}}

Solutions:

1.

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2. 16^{\dfrac{1}{4}}=\left( 2^{4}\right) ^{\dfrac{1}{4}}=2^{1}=2

3. 125^{-\dfrac{2}{3}}=\left( 5^{3}\right) ^{-\dfrac{2}{3}}=5^{-2}=\dfrac{1}{5^{2}}=\dfrac{1}{25}

4. 64^{-\dfrac{4}{3}}=\left( 4^{3}\right) ^{-\dfrac{4}{3}}=4^{-4}=\dfrac{1}{4^{4}}=\dfrac{1}{256}

5. 36^{-\dfrac{1}{2}}=\left( 6^{2}\right) ^{-\dfrac{1}{2}}=6^{-1}=\dfrac{1}{6}

6. 81^{\dfrac{5}{4}}=\left( 3^{4}\right) ^{\dfrac{5}{4}}=3^{5}=243

7. \left( \dfrac{2}{3}\right) ^{-2}=\dfrac{1}{\left( \dfrac{2}{3}\right) ^{2}}=\dfrac{1}{\dfrac{4}{9}}=1\div \dfrac{4}{9}=1\times \dfrac{9}{4}=\dfrac{9}{4}

  • Note that, \left( \dfrac{a}{b}\right) ^{-m}=\left( \dfrac{b}{a}\right) ^{n}

8. -8^{\dfrac{2}{3}}=\left[ \left( -2\right) ^{3}\right] ^{\dfrac{2}{3}}=\left( -2\right) ^{2}=4

9. -27^{-\dfrac{4}{3}}=\left[ \left( -3\right) ^{3}\right] ^{-\dfrac{4}{3}}=\left( -3\right) ^{-4}=\dfrac{1}{\left( -3\right) ^{4}}=\dfrac{1}{81}


Example

Evaluate:

1. { 16 }^{ \frac { 1 }{ 4 } }=({ { 2 }^{ 4 }) }^{ \frac { 1 }{ 4 } }={ 2 }^{ 1 }=2

  • We first write it as 2^4, then apply the power.

2. { 16 }^{ -\frac { 3 }{ 4 } }=({ { 2 }^{ 4 }) }^{ \frac { 3 }{ 4 } }={ 2 }^{ -2 }=\frac { 1 }{ { 2 }^{ 3 } } =\frac { 1 }{ 8 }

3. { 125 }^{ -\frac { 1 }{ 3 } }=({ { 5 }^{ 3 }) }^{ -\frac { 1 }{ 3 } }={ 5 }^{ -1 }=\frac { 1 }{ { 5 }^{ 1 } } =\frac { 1 }{ 5 }

4. ({ \frac { 9 }{ 16 } })^{ -\frac { 3 }{ 2 } }=(({ { \frac { 3 }{ 4 } ) }^{ 2 }) }^{ -\frac { 3 }{ 2 } }={ (\frac { 3 }{ 4 } })^{ -3 }=\frac { 1 }{ { (\frac { 3 }{ 4 } ) }^{ 3 } }

  • =\frac { 1 }{ \frac { 27 }{ 64 } }
  • =\frac { 64 }{ 27 }

5. ({ 2\frac { 1 }{ 4 } })^{ 3 }={ (\frac { 9 }{ 4 } ) }^{ 3 }=\frac { 729 }{ 64 }

6. ({ 2\frac { 1 }{ 4 } ) }^{ -3 }=({ \frac { 9 }{ 4 } ) }^{ -3 }=({ \frac { 4 }{ 9 } ) }^{ 3 }=\frac { 64 }{ 729 }

7. ({ 2\frac { 1 }{ 4 } ) }^{ \frac { 3 }{ 2 } }=({ \frac { 9 }{ 4 } ) }^{ \frac { 3 }{ 2 } }=(({ { \frac { 3 }{ 2 } ) }^{ 2 } })^{ \frac { 3 }{ 2 } }={ (\frac { 3 }{ 2 } ) }^{ 3 }=\frac { 27 }{ 8 }

8. ({ 2\frac { 1 }{ 4 } ) }^{ -\frac { 3 }{ 2 } }=({ \frac { 9 }{ 4 } ) }^{ -\frac { 3 }{ 2 } }=(({ { \frac { 3 }{ 2 } ) }^{ 2 }) }^{ -\frac { 3 }{ 2 } }

  • ={ (\frac { 3 }{ 2 } ) }^{ -3 }
  • =({ \frac { 2 }{ 3 } ) }^{ 3 }
  • =\frac { 8 }{ 27 }

9. { 25 }^{ -\frac { 1 }{ 2 } }=({ { 5 }^{ 2 }) }^{ -\frac { 1 }{ 2 } }={ 5 }^{ -1 }=\frac { 1 }{ 5 }

10. { 49 }^{ -\frac { 3 }{ 2 } }=({ { 7 }^{ 2 }) }^{ -\frac { 3 }{ 2 } }={ 7 }^{ -3 }=\frac { 1 }{ { 7 }^{ 3 } } =\frac { 1 }{ 243 }

  • Note { a }^{ \frac { 1 }{ a } }=\sqrt [ n ]{ a } and { a }^{ \frac { m }{ n } }=\sqrt [ n ]{ { a }^{ m } } so 16^{ \frac { 1 }{ 2 } }=\sqrt { 16 } and \sqrt [ 3 ]{ 81 } ={ 81 }^{ \frac { 1 }{ 4 } }=3

11. \sqrt [ 3 ]{ { 125 }^{ 2 } } =({ { 125 }^{ 2 }) }^{ \frac { 1 }{ 3 } }={ 125 }^{ \frac { 2 }{ 3 } }={ 5 }^{ 3\times \frac { 2 }{ 3 } }={ 5 }^{ 2 }=25

12. \sqrt [ 3 ]{ { 343 }^{ -1 } } =({ { 343 }^{ -1 }) }^{ \frac { 1 }{ 3 } }=({ { 7 }^{ 3 }) }^{ -\frac { 1 }{ 3 } }={ 7 }^{ -1 }=\frac { 1 }{ 7 }

13. \sqrt { 25 } +\sqrt { 36 } +\sqrt { 49 } =5+6+7=18

14. \sqrt { 0.0064 } =\sqrt { \frac { 64 }{ 10000 } } =\frac { \sqrt { 64 } }{ \sqrt { 10000 } } =\frac { 8 }{ 100 } =0.08

15. \frac { { 2 }^{ \frac { 1 }{ 3 } }\times { 2 }^{ \frac { 4 }{ 3 } } }{ { 2 }^{ \frac { 2 }{ 3 } } }

  • Using the laws of indices, we know that { a }^{ n }\times { a }^{ m }={ a }^{ n+m } and { a }^{ n }\div { a }^{ m }={ a }^{ n-m }. So:
  • \frac { { 2 }^{ \frac { 1 }{ 3 } }\times { 2 }^{ \frac { 4 }{ 3 } } }{ { 2 }^{ \frac { 2 }{ 3 } } } ={ 2 }^{ \frac { 1 }{ 3 } +\frac { 4 }{ 3 } -\frac { 2 }{ 3 } }={ 2 }^{ \frac { 3 }{ 3 } }={ 2 }^{ 1 }=2

16. { 8 }^{ \frac { 2 }{ 3 } }\times { 8 }^{ -\frac { 1 }{ 3 } }={ 8 }^{ \frac { 2 }{ 3 } +(-3) }={ 8 }^{ \frac { 2 }{ 3 } -\frac { 1 }{ 3 } }={ 8 }^{ \frac { 1 }{ 3 } }=({ { 2 }^{ 3 }) }^{ \frac { 1 }{ 3 } }={ 2 }^{ 1 }=2


Example

This example is from paper 1 of a Cambridge exam, from Nov/Oct 2002.

Evaluate:

1. { 7 }^{ \frac { 1 }{ 4 } }-{ 6 }^{ \frac { 2 }{ 3 } }

2. 3\sqrt { 2 } \times 5\sqrt { 2 }

Solution:

1. 7\frac { 1 }{ 4 } =\frac { 29 }{ 4 } and 6\frac { 2 }{ 3 } =\frac { 20 }{ 3 }

Therefore, 7\frac { 1 }{ 4 } -6\frac { 2 }{ 3 } =\frac { 29 }{ 4 } -\frac { 20 }{ 5 } =\frac { 29\times 3 }{ 4\times 3 } -\frac { 20\times 4 }{ 3\times 4 }

=\frac { 87-80 }{ 12 }

=\frac { 7 }{ 12 }

2. 3\sqrt { 2 } \times 5\sqrt { 2 } =3\times 5\times \sqrt { 2 } \times \sqrt { 2 }

15\times ({ \sqrt { 2 } ) }^{ 2 }

15\times 2

=30