# 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.   $27 \left(\dfrac{1}{9}\right)^2 \left(\dfrac{9^2}{3^5} \right) =$ 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) $= \dfrac{1}{3^2} = \dfrac{1}{9}$ $80 \left(\dfrac{1}{5^{-1}}\right)^2 \left(\dfrac{25^{-1}}{4}\right)^2 =$ 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 $= \dfrac {1}{5}$ 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 $= \dfrac{x^{12-4}}{ y^{15-4} } = \dfrac{x^8}{ y^{11}}$ 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 $= 24 (x + y)^2$ 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 $= \dfrac{y^6}{4 x^6}$ $(- 3 x^2 y^3) (- 4 x^3 y^5) =$ 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 $= 12 x^5 y^8$ 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 $= 3 \dfrac{x^5}{ y^5}$ 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 $= \dfrac{x^4 y^4}{ 9}$ 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 $\dfrac{5}{16} x y^7$ 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 $= \dfrac{x^{6-2}}{y^9}$ $= \dfrac{x^4}{y^9}$ 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 $= \dfrac{1}{2} \dfrac{x^7}{y^4}$