# 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
$$= 36 a^2$$