 # 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.