# 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
$= \dfrac{1}{2} \dfrac{x^7}{y^4}$

1. E
2. A
3. B
4. C
5. D
6. B
7. A
8. E
9. C
10. D
11. B