Solutions to Exponents Questions for Grade 9

Detailed solutions to Grade 9 questions on exponents are presented.
The exponent rules to be used to answer the questions below are included.

Solutions to Questions

  1. Evaluate the following.

    1. ) \( 1^1 = 1\)

    2. ) \( 2^3 = 2 \cdot 2 \cdot 2 = 8\)

    3. ) \( (-2)^2 = (-2) \cdot (-2) = 4\)

    4. ) \( (-2)^3 = (-2) \cdot (-2) \cdot (-2) = - 8 \)

    5. ) \( 3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81\)

    6. ) \( 4^2 = 4 \cdot 4 = 16 \)

    7. ) \( 2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32\)

    8. ) \( 5^2 = 5 \cdot 5 = 25 \)

    9. ) \( (-1)^6 = (-1) \cdot (-1) \cdot (-1) \cdot (-1) \cdot (-1) \cdot (-1) = 1\)

    10. ) \( 7^2 = 7 \cdot 7 = 49 \)

    11. ) \( (-9)^2 = (-9) \cdot (-9) = 81\)

    12. ) \( 3^3 = 3 \cdot 3 \cdot 3 = 27 \)

    13. ) \( 10^2 = 10 \cdot 10 = 100\)

    14. ) \( 10^3 = 10 \cdot 10 \cdot 10 = 1000\)

    15. ) \( 0.1^3 = 0.1 \cdot 0.1 \cdot 0.1 = 0.001\)


  2. Write the following numbers in exponent form with exponent not equal to \( 1 \). There might be more than one answer.

    1. ) \( 0 = 0^1 = 0^3 = 0^5 ...\) many possible answers (NOTE: \( 0^0 \) is undefined).

    2. ) \( 1 = 1^0 = 1^1 = 1^2 = ... \) many possible answers

    3. ) \( 4 = 2^2 = (-2)^2\)

    4. ) \( 8 = 2^3\)

    5. ) \( 9 = 3^2 = (-3)^2\)

    6. ) \( 16 = 2^4 = 4^2 = (-2)^4 = (-4)^2\)

    7. ) \( 25 = 5^2 = (-5)^2\)

    8. ) \( 32 = 2^5 \)

    9. ) \( 49 = 7^2 = (-7)^2\)

    10. ) \( 64 = 8^2 = 4^3 = (-8)^2 \)

    11. ) \( 81 = 9^2 = (-9)^2 \)

    12. ) \( 100 = 10^2 = (-10)^2\)

    13. ) \( -27 = (-3)^3\)

    14. ) \( -8 = (-2)^3 \)

    15. ) \( -64 = (-4)^3 \)


  3. Use the above rules to evaluate the following expressions. Some solutions are presented but there other ways to answer these questions.

    1. ) \( 120^0 = 1\)     rule 9

    2. ) \( 2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8} = 0.125 \)     rule 2

    3. ) \( 2^{-3} \cdot 2^6 = 2^{-3+6} = 2^{3} = 8\)     rule 3

    4. ) \( 2^{3} \cdot 3^3 = (2 \cdot 3)^3 = 6^3 = 216\)     rule 4

    5. ) \( \dfrac{3^{10}}{3^8} = 3^{10-8} = 3^2 = 9 \)     rule 5

    6. ) \( 4^{-1} = \dfrac{1}{4^1} = \dfrac{1}{4} = 0.25\)     rule 2

    7. ) \( \dfrac{8^3}{4^3} = \left(\dfrac{8}{4} \right)^3 = 2^3 = 8\)     rule 6

    8. ) \( \dfrac{100^3}{10^3} = \left(\dfrac{100}{10} \right)^3 = 10^3 = 1000\)     rule 6

    9. ) \( (2^{2})^2 = 2^{2 \cdot 2} = 2^4 = 16\)     rule 8

    10. ) \( (1^{3})^{25} = 1^{3 \cdot 25} = 1^{75} = 1 \)     rule 8 and 11

    11. ) \( ((-1)^{2})^{20} = (-1)^{2 \cdot 20} = (-1)^{40} = 1 \)     rule 8 and 12

    12. ) \( - 2^{-2} = - \dfrac{1}{2^2} = - \dfrac{1}{4} = - 0.25\)     rule 2

    13. ) \( ((-1)^{-1})^{-1} = (-1)^{(-1) \cdot (-1)} = (-1)^1 = - 1 \)     rule 8 and 12

    14. ) \( \left (\dfrac{100}{10} \right)^{-2} = \left (\dfrac{10}{100} \right)^{2} = 0.1^2 = 0.01\)     rule 7

    15. ) \( \left (\dfrac{10}{1000} \right)^{-2} = \left (\dfrac{1000}{10} \right)^{2} = 100^2 = 10000 \)     rule 7


  4. Simplify and write the following expressions with a single positive exponent if possible.

    1. ) \( 3^2 \cdot 3^8 = 3^{2+8} = 3^{10} \)     rule 3

    2. ) \( \dfrac{2^5}{2^2} = 2^{5-2} = 2^3\)     rule 5

    3. ) \( \left( {3^5} \right)^2 = 3^{5\cdot2} = 3^{10}\)     rule 8

    4. ) \( 6^4 \cdot \dfrac{6^5}{6^2} = 6^4 \cdot 6^{5-2} = 6^4 \cdot 6^{3} = 6^{4+3} = 6^7 \)     rules 5 and 3

    5. ) \( (-7)^2 \cdot (-7)^3 = (-7)^{2+3} = (-7)^5\)     rule 3

    6. ) \( \left( {5^2} \right)^2 \cdot \left( {5^3} \right)^3 \cdot 5 = \left( {5^2} \right)^2 \cdot \left( {5^3} \right)^3 \cdot 5^1 =5^{2+3+1} = 5^6\)     rule 3 extended to 3 terms

    7. ) \( x^{-1} x^3 = x^{-1+3} = x^2\)     rule 3

    8. ) \( \dfrac{a^5}{a^2} = a^{5-2} = a^3\)     rule 5

    9. ) \( \dfrac{a^2}{a^7} = a^{2-7} = a^{-5} = \dfrac{1}{a^5} \)     rule 5 and 2

    10. ) \( 2^{x} \cdot 4^3 \cdot 2^y \)
      Write \( 4 \) as \( 2^2 \) and rewrite the given expression as
      \( = 2^{x} \cdot (2^2)^3 \cdot 2^y \)
      Use rule 8 to rewrite \( (2^2)^3 \) as \( 2^6 \)
      \( = 2^{x} \cdot 2^6 \cdot 2^y \)
      Use rule 3 extended to three terms with the same base
      \( = 2^{x + 6 + y} \)

    11. ) \( (3^{-1})^x \)
      Use rule 8 to rewrite the given expression as
      \( = 3^{(-1) \cdot x} \)
      Simplify exponent: \( (-1) \cdot x = - x \)
      \( = 3^{-x} \)
      Use rule 2 to rewrite as
      \( = \dfrac{1}{3^x} \)

    12. ) \( 3^{x} \cdot 9^{x} \)
      Use rule 4
      \( = (3 \cdot 9)^x\)
      Simplify
      \( = 27^x \)

    13. ) \( \dfrac{a^x}{a^4} a^6 \)
      Use rule 5 to write \( \dfrac{a^x}{a^4} \) as \( a^{x-4} \) and substitute in the given expression
      \( = a^{x-4} \cdot a^6 \)
      Use rule 3
      \( = a^{x-4 + 6} \)
      Simplify
      \( = a^{x+2} \)


  5. Simplify following expressions.

    1. ) \( a^2 \cdot \dfrac{a^5}{a^2} = a^2 \cdot a^{5-2} = a^2 \cdot a^3 = a^{2+3} = a^5 \)

    2. ) \( \left (\dfrac{3 x}{x} \right)^3\)
      Simplify by cancelling \( x \) inside the bracket
      \( = \left (\dfrac{3 }{ 1} \right)^3 = 3^3 = 27\)

    3. ) \( (2^{2})^2 \) Evaluate \( 2^{2} \) to \( 4 \) inside the bracket
      \( = 4^2 = 16 \)

    4. ) \( \dfrac{1}{4} \cdot \left (\dfrac{2x}{x} \right)^2\) Simplify by cancelling \( x \) inside the bracket
      \( = \dfrac{1}{4} \cdot \left (\dfrac{2}{1} \right)^2\)
      \( = \dfrac{1}{4} \cdot 2^2 = \dfrac{1}{4} \cdot 4 = 1\)

    5. ) \( \dfrac{y^4 x^3}{x^2 y^2} \)
      Rewrite as the product of two fractions where the numerator and denominator of each fraction has the same variable
      \( = \dfrac{x^3}{x^2} \cdot \dfrac{y^4}{y^2} \)
      Apply rule 5 to each fraction
      \( = x^{3-2} \cdot y^{4-2} \)
      Simplify
      \( = x y^2 \)

    6. ) \( \dfrac{x^2}{4y^2} \cdot \left (\dfrac{8y}{x} \right)^2\)
      Apply rule 6 to the term \( \left (\dfrac{8y}{x} \right)^2\)
      \( = \dfrac{x^2}{4y^2} \cdot \dfrac{(8y)^2}{x^2} \)
      Apply rule 4 to the term \( (8y)^2 \)
      \( = \dfrac{x^2}{4y^2} \cdot \dfrac{8^2 y^2}{x^2} \)
      Rewrite as the product of three fractions
      \( = \left(\dfrac{8^2}{4} \cdot \right) \left(\dfrac{x^2}{x^2}\right) \cdot \left(\dfrac{y^2}{y^2}\right) \)

      Rewrite \( 8^2 \) as \( (2^3)^2 = 2^6 \) and \( 4 \) as \( 2^2 \)

      \( = \left(\dfrac{2^6}{2^2} \cdot \right) \left(\dfrac{x^2}{x^2}\right) \cdot \left(\dfrac{y^2}{y^2}\right) \)

      Use rule 5 to the fraction \( \dfrac{2^6}{2^2} \) and cancel same terms in the other two fractions

      \( = 2^{6-2} \cdot 1 \cdot 1 = 2^4 = 16 \)


    7. ) \( ( - 6 a)^2 \cdot (a^2 + 1)^0 \)
      Use rule 9 to simplify \( (a^2 + 1)^0 \) to 1
      \( = ( - 6 a)^2 \cdot 1 \)
      Use rule 4 extended to the three terms in \( ( - 6 a)^2 \)
      \( = (-1)^2 6^2 a^2 \)
      Simplify
      \( = 36 a^2 \)


Solutions and detailed explanations to the above questions.

More References and Links

Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers
Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers Home Page

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