Solutions to Questions on Simplifying Exponents


Solutions with detailed explanations to Questions on Simplifying Exponents are presented.
The rules of of exponents are applied to solve the following questions.
    \( \) \( \)\( \)\( \)


  1. \( 27 \left(\dfrac{1}{9}\right)^2 \left(\dfrac{9^2}{3^5} \right) = \)
    Solution
    Use the quotient rule of exponents \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to rewrite \( \left(\dfrac{1}{9}\right)^2 \) in the given expression as
    \( 27 \left(\dfrac{1}{9}\right)^2 \left(\dfrac{9^2}{3^5} \right) = 27 \; \dfrac{1^2}{9^2} \; \dfrac{9^2}{3^5} \)
    Rewrite \( 27 \) as \( 3^3 \)
    \( 3^3 \; \dfrac{1^2}{9^2} \; \dfrac{9^2}{3^5} \)
    Simplify \( 1^2 = 1 \), remove brackets, multiply and rewrite as one fraction,
    \( = \dfrac{3^3 9^2}{9^2 3^5} \)
    Divide the numerator and the denominator by the GCF \(= 3^3 9^2\) of the two and simplify (or cancel common factors)
    \( = \dfrac{1}{3^2} = \dfrac{1}{9} \)


  2. \( 80 \left(\dfrac{1}{5^{-1}}\right)^2 \left(\dfrac{25^{-1}}{4}\right)^2 = \)
    Solution
    Use rules of exponents \( a^m \cdot b^m = (a \cdot b)^m \)
    to rewrite \( \left(\dfrac{1}{5^{-1}}\right)^2 \left(\dfrac{25^{-1}}{4}\right)^2 \) as \( \left(\dfrac{1}{5^{-1}} \dfrac{25^{-1}} {4} \right)^2 \) in the given the expression
    \( 80 \left(\dfrac{1}{5^{-1}}\right)^2 \left(\dfrac{25^{-1}}{4}\right)^2 = 80 \left(\dfrac{1}{5^{-1}} \dfrac{25^{-1}} {4} \right)^2 \)
    Change negative exponents to positive exponents using the rule \( a^{-n} = \dfrac{1}{a^n} \) to rewrite \( 5^{-1} = \dfrac {1}{5} \) and \( 25^{-1} = \dfrac{1}{25} \)
    \( = 80 \left(\dfrac{1}{\dfrac{1}{5}} \dfrac{\dfrac{1}{25}} {4} \right)^2 \)
    Use division of fractions to rewrite \( \dfrac{1}{\dfrac{1}{5}} = 5 \) and \( \dfrac{\dfrac{1}{25}} {4} = \dfrac{1}{25 \times 4} = \dfrac{1}{100} \) to rewrite the expression as
    \( = 80 \left( 5 \times \dfrac{1}{100} \right)^2\)
    Multiply inside brackets
    \( = 80 \left( \dfrac{5}{100} \right)^2\)
    Divide numerator and denominator by \( 5 \) and simplify
    \( = 80 \left( \dfrac{1}{20} \right)^2\)
    Use the rule of exponents \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to rewrite \( \left( \dfrac{1}{20} \right)^2\) in the given expression
    \( = 80 \times \dfrac{1^2}{20^2} \)
    Simplify
    \( = \dfrac{80}{400} \)
    Divide numerator and denominator by the common factor \( 80 \) to simplify the given expression to
    \( = \dfrac {1}{5} \)


  3. For x and y not equal to zero, the expression
    \( \left(\dfrac{x^4}{y^5}\right)^3 \left(\dfrac{y^2}{x^2}\right)^2 \)
    simplifies to
    Solution
    Use the rule of exponents \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) and remove brackets to rewrite the given expression as
    \( \left(\dfrac{x^4}{y^5}\right)^3 \left(\dfrac{y^2}{x^2}\right)^2 = \dfrac{(x^4)^3}{(y^5)^3} \dfrac{(y^2)^2}{(x^2)^2} \)
    Use the rule of exponents \( (a^n)^m = a^{n \cdot m} \) to rewrite the expression as
    \( = \dfrac{x^{12}}{y^{15}} \dfrac{y^4}{x^4} \)
    Apply the quotient rule \( \dfrac{a^m}{a^n} = a^{m - n} \) to simplify
    \( = \dfrac{x^{12-4}}{ y^{15-4} } = \dfrac{x^8}{ y^{11}}\)


  4. For \( x \ne - y \), \( \dfrac{3(2x + 2y)^5}{4 (x + y)^3} = \)
    Solution
    Factor \( 2 \) out in the expression \( 2x + 2y \) and write given expression as
    \( \dfrac{3(2x + 2y)^5}{4 (x + y)^3} = \dfrac{3(2(x + 2)y)^5}{4 (x + y)^3} \)
    Apply exponent rule \( (a \cdot b)^m = a^m \cdot b^m \) to write expression as
    \( = \dfrac{3 \times 2^5(x + 2)^5}{4 (x + y)^3} \)
    Rearrange as
    \( = \dfrac{3 \times 2^5}{2^4} \dfrac{(x + y)^5}{(x + y)^3} \)
    Use the quotient rule of exponents \( \dfrac{a^m}{a^n} = a^{m - n} \) to rewrite the above as
    \( = 3 \times 2^{5-4} \times (x + y)^{5-3} \)
    Simplify
    \( = 24 (x + y)^2 \)


  5. For x and y not equal to zero, the expression
    \( \left(\dfrac{x^0}{2y}\right)^2 \left(\dfrac{y^4}{x^3}\right)^2 \)
    simplifies to
    Solution
    Use rules of exponents \( x^0 = 1 \) and \( a^m \cdot b^m = (a \cdot b)^m \) to rewrite given expression as
    \( \left(\dfrac{x^0}{2y}\right)^2 \left(\dfrac{y^4}{x^3}\right)^2 = \left(\dfrac{1 \times y^4}{2y \times x^3}\right)^2 \)
    Simplify inside brackets
    \( = \left(\dfrac{y^{4-1}}{2 x^3}\right)^2 = \left(\dfrac{y^3}{2 x^3}\right)^2\)
    Apply \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to rewrite the above as
    \( = \dfrac{y^6}{4 x^6} \)


  6. \( (- 3 x^2 y^3) (- 4 x^3 y^5) = \)
    Solution
    Multiply and put terms with the same variable within brackets
    \( (- 3 x^2 y^3) (- 4 x^3 y^5) = (-3 \times (-4)) (x^2 \times x^3)(y^3 \times y^5) \)
    Apply the exponent rule \( a^m \cdot a^n = a^{m+n} \)
    \( = 12 (x^{2+3})(y^{3+5}) \)
    Simplify
    \( = 12 x^5 y^8 \)


  7. For x and y not equal to zero, the expression
    \( \dfrac{12 x^3 y^{-2}}{4 x^{-2}y^3} \)
    simplifies to
    Solution
    Rearrange and write as the product of rational expressions with the same variable
    \( \dfrac{12 x^3 y^{-2}}{4 x^{-2}y^3} = \dfrac{12}{4} \times \dfrac{x^3}{x^{-2}} \times \dfrac{y^{-2}}{y^3} \)
    Apply the exponent rule \( \dfrac{a^m}{a^n} = a^{m - n} \)
    \( = 3 x^{3-(-2)} )(y^{-2-3}) \)
    Simplify
    \( = 3 x^5 y^{-5} \)
    Write with positive integer \( = 3 \dfrac{x^5}{ y^5} \)


  8. For x and y not equal to zero,
    \( \left(\dfrac{3 x^{-2}}{y^2}\right)^{-2} = \)
    For x and y not equal to zero,
    Solution
    Apply the exponent rule \( \left( \dfrac{a}{b} \right)^{-m} = \dfrac{b^m}{a^m} \) to rewrite the given expression as
    \( \left(\dfrac{3 x^{-2}}{y^2}\right)^{-2} = \dfrac{({y^2})^2}{(3 x^{-2})^2} \)
    Apply the rule exponents \( (a^n)^m = a^{n \cdot m} \) in the numerator and the exponent rule \( (a \cdot b)^m = a^m \cdot b^m \) in the denominator
    \( = \dfrac{y^4}{3^2 (x^{-2})^2} \)
    Simplify
    \( = \dfrac{y^4}{ 9 x^{-4}} \)
    Write with positive exponents
    \( = \dfrac{x^4 y^4}{ 9} \)


  9. For x and y not equal to zero, the expression
    \( \left( \dfrac{2x^2 y^{-1}}{5} \right)^2 \left(\dfrac{5 x^{-1}y^3}{4}\right)^3 \)
    simplifies to
    Solution
    Apply the exponent rule \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to rewrite the given expression as
    \( \left( \dfrac{2x^2 y^{-1}}{5} \right)^2 \left(\dfrac{5 x^{-1}y^3}{4}\right)^3 = \dfrac{(2x^2 y^{-1})^2}{5^2} \dfrac{(5 x^{-1}y^3)^3}{4^3} \)
    Apply the exponent rule \( (a \cdot b)^m = a^m \cdot b^m \) in the numerators
    \( = \dfrac{2^2 (x^2)^2 (y^{-1})^2 }{5^2} \dfrac{5^3 (x^{-1})^3 (y^3)^3}{4^3} \)
    Apply the rule exponents \( (a^n)^m = a^{n \cdot m} \) in the numerator to rewrite the above as
    \( = \dfrac{2^2 x^4 y^{-2} }{5^2} \dfrac{5^3 x^{-3} y^9}{4^3} \)
    Write with positive exponents using the rule of negative exponents \( a^{-n} = \dfrac{1}{a^n} \)
    \( = \dfrac{2^2 x^4 }{5^2} \dfrac{1}{y^2} \dfrac{5^3 y^9}{4^3} \dfrac{1}{x^3} \)
    Multiply
    \( = \dfrac{2^2 x^4 }{5^2 y^2} \dfrac{5^3 y^9}{4^3 x^3} \)
    Rewrite as a multiplication of rational (fractional) expressions of the same variable
    \( = \dfrac{2^2 x^4 }{5^2 y^2} \dfrac{5^3 y^9}{4^3 x^3} = \dfrac{4 \times 5^3}{4^3 \times 5^2} \dfrac{x^4}{x^3} \dfrac{y^9}{y^2} \)
    Apply the rule of exponents \( \dfrac{a^m}{a^n} = a^{m - n} \)
    \( \dfrac{5^{3-2}}{4^{3-1}} x^{4-3} y^{9-2} \)
    Simplify
    \( \dfrac{5}{16} x y^7 \)


  10. For x and y not equal to zero,
    \( \left(\dfrac{x^{-1}}{y^0} \right)^2 \left(\dfrac{x^2}{y^3}\right)^3 = \)
    Solution
    Apply the rule of exponents \( y^0 = 1 \) and the rule \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) and simplify
    \( \left(\dfrac{x^{-1}}{y^0} \right)^2 \left(\dfrac{x^2}{y^3}\right)^3 = (x^{-1})^2 \dfrac{(x^2)^3}{(y^3)^3} \)
    Apply the rule exponents \( (a^n)^m = a^{n \cdot m} \) to rewrite the above as
    \( = x^{-2} \dfrac{x^6}{y^9} \)
    Multiply and simplify
    \( = \dfrac{x^{6-2}}{y^9} \)
    \( = \dfrac{x^4}{y^9} \)


  11. For \( y \ne 0 \), the expression
    \( \left(\dfrac{x^2}{y^3}\right) \left(\dfrac{8x}{y}\right) \left(\dfrac{x^2}{4}\right)^2 \)
    simplifies to
    Solution
    Apply the rule \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to the term on the right
    \( \left(\dfrac{x^2}{y^3}\right) \left(\dfrac{8x}{y}\right) \left(\dfrac{x^2}{4}\right)^2 = \left(\dfrac{x^2}{y^3}\right) \left(\dfrac{8x}{y}\right) \left(\dfrac{x^4}{4^2}\right) \)
    Multiply
    \( = \dfrac{x^2 \times 8x \times x^4}{y^3 \times y \times 16} \)
    Apply the rule \( a^m \cdot a^n = a^{m+n} \)
    \( = \dfrac{8 x^{2+1+4}}{16 y^{3+1}} \)
    Simplify
    \( = \dfrac{1}{2} \dfrac{x^7}{y^4} \)

Answers to the Above Questions
  1. E
  2. A
  3. B
  4. C
  5. D
  6. B
  7. A
  8. E
  9. C
  10. D
  11. B



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