PSEB 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.2

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 13 Exponents and Powers Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2

1. Using laws of exponents, simplify and write the following in the exponential form :

(i) 2⁷ × 2⁴
(ii) p⁵ × p³
(iii) (-7)⁵ × (-7)¹¹
(iv) 20¹⁵ ÷ 20¹³
(v) (-6)⁷ ÷ (-6)³
(vi) 7^x × 7³
Solution:
(i) 2⁷ × 2⁴ = 2⁷⁺⁴ = 211
(ii) p⁵ × p³ = p⁵⁺³ = p8
(iii) (-7)⁵ × (-7)¹¹ = (-7)⁵⁺¹¹ = (-7)¹⁶
(iv) 20¹⁵ ÷ 20¹³ = 20^15-13 = 20²
(v) (-6)⁷ ÷ (-6)³ = (-6)^7-3 = (-6)⁴
(vi) 7^x × 7³ = 7^x+3

2. Simplify and write the following in exponential form.

(i) 5³ × 5⁷ × 5¹²
Solution:
5³ × 5⁷ × 5¹² = 5³⁺⁷⁺¹²
= 5²²

(ii) a⁵ × a³ × a⁷
Solution:
a⁵ × a³ × a⁷ = a⁵⁺³⁺⁷
= a¹⁵

3. Simplify and write the following in the exponential form :

(i) (2²)¹⁰⁰
Solution:
(2²)100 = 2^{2 × 100}
= 2²⁰⁰

(ii) (5³)⁷
Solution:
(5³)⁷ = 5^{3 × 7}
= 5²¹

4. Simplify and write in the exponential form:

(i) (2³)⁴ ÷ 2⁵
Solution:
(2³)⁴ ÷ 2⁵ = 2¹² ÷ 2⁵
= 2^12-5
= 2⁷

(ii) 2³ × 2² × 5⁵
Solution:
2³ × 2² × 5⁵ = 2³⁺² × 5⁵
= 2⁵ × 5⁵
= (2 × 5)⁵
= 10⁵

(iii) [(22)³ × 3⁶] × 5⁶
Solution:
[(2²)³ × 3⁶] × 5⁶ = [2^2×3 × 3⁶] × 5⁶
= [2⁶ × 3⁶] × 5⁶
= 6⁶ × 5⁶
= (6 × 5)⁶
= 30⁶.

5. Simplify and write in the exponential form:

(i) 5⁴ × 8⁴
Solution:
5⁴ × 8⁴ = (5 × 8)⁴
= 40⁴

(ii) (-3)⁶ × (-5)⁶
Solution:
(-3)⁶ × (-5)⁶ = (-3 × -5)⁶
= (+15)⁶

6. Simplify and express each of the following in the exponential form :

(i) \(\frac{\left(3^{2}\right)^{3} \times(-2)^{5}}{(-2)^{3}}\)
Solution:
\(\frac{\left(3^{2}\right)^{3} \times(-2)^{5}}{(-2)^{3}}=\frac{3^{2 \times 3} \times(-2)^{5}}{(-2)^{3}}\)
= 3⁶ × (-2)^5-3
= 3⁶ × (-2)²
= 3⁶ × 2²

(ii) \(\frac{3^{7}}{3^{4} \times 3^{3}}\)
Solution:
\(\frac{3^{7}}{3^{4} \times 3^{3}}=\frac{3^{7}}{3^{4+3}}=\frac{3^{7}}{3^{7}}\)
= 3^7-7
= 3⁰
= 1¹

(iii) \(\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}\)
Solution:
\(\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}=\frac{2^{8}}{\left(2^{2}\right)^{3}} \times \frac{a^{5}}{a^{3}}\)
= \(\frac{2^{8}}{2^{6}} \times a^{5-3}\)
= \(2^{8-6} \times a^{5-3}\)
= \(2^{2} \times a^{2}\)
= (2a)²

(iv) 3⁰ × 4⁰ × 5⁰
Solution:
3⁰ × 4⁰ × 5⁰
= 1 × 1 × 1
= 1

7. Express each of the following rational number in the exponontial form :

(i) \(\frac {25}{64}\)
Solution:
\(\frac {25}{64}\) = \(\frac{5 \times 5}{8 \times 8}=\frac{5^{2}}{8^{2}}\)
= \(\left(\frac{5}{8}\right)^{2}\)

(ii) \(\frac {-64}{125}\)
Solution:
\(\frac {-64}{125}\) = \(\frac{-4 \times 4 \times 4}{5 \times 5 \times 5}\)
= \(\frac{(-4)^{3}}{5^{3}}\)
= \(\left(-\frac{4}{5}\right)^{3}\)

(iii) \(\frac {-125}{216}\)
Solution:
\(\frac {-125}{216}\) = \(\frac{-5 \times 5 \times 5}{6 \times 6 \times 6}\)
= \(\frac{(-5)^{3}}{6^{3}}\)
= \(\left(-\frac{5}{6}\right)^{3}\)

(iv) \(\frac {-343}{729}\)
Solution:
\(\frac {-343}{729}\) = \(\frac{-7 \times 7 \times 7}{9 \times 9 \times 9}\)
= \(\frac{(-7)^{3}}{9^{3}}\)
= \(\left(-\frac{7}{9}\right)^{3}\)

8. Simplify :

(i) \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\)
Solution:
\(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\) = \(\frac{2^{5 \times 2} \times 7^{3}}{\left(2^{3}\right)^{3} \times 7}\)
= \(\frac{2^{10} \times 7^{3}}{2^{9} \times 7}\)
= 2^10-9 × 7^3-1
= 2¹ × 7²
= 2 × 7 × 7
= 98

(ii) \(\frac{2 \times 3^{4} \times 2^{5}}{9 \times 4^{2}}\)
Solution:
\(\frac{2 \times 3^{4} \times 2^{5}}{9 \times 4^{2}}\) = \(\frac{2 \times 2^{5} \times 3^{4}}{3 \times 3 \times\left(2^{2}\right)^{2}}\)
= \(\frac{2^{1+5} \times 3^{4}}{3^{2} \times 2^{4}}\)
= 2^6-4 × 3^4-2
= 2² × 3²
= 2 × 2 × 3 × 3
= 36.

9. Express each of the following as a product of prime factors in the exponential form

(i) 384 × 147
Solution:
384 × 147
384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 2⁷ × 3¹
147 = 7 × 7 × 3
= 7² × 3¹
\(\begin{array}{l|l}
2 & 384 \\
\hline 2 & 192 \\
\hline 2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

\(\begin{array}{l|l}
7 & 147 \\
\hline 7 & 21 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

384 × 147 = 2⁷ × 3¹ × 7² × 3¹
= 2⁷ × 3² × 7²

(ii) 729 × 64
Solution:
729 × 64
729 = 3 × 3 × 3 × 3 × 3 × 3
= 3⁶
\(\begin{array}{l|l}
3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
64 = 2 × 2 × 2 × 2 × 2 × 2
= 2⁶
\(\begin{array}{l|l}
2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
= 729 × 64 = 3⁶ × 2⁶

(iii) 108 × 92
Solution:
108 = 2 × 2 × 3 × 3 × 3
= 2² × 3³
\(\begin{array}{c|c}
2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
92 = 2 × 2 × 23
= 2² × 23
\(\begin{array}{l|l}
2 & 92 \\
\hline 2 & 46 \\
\hline & 23
\end{array}\)
108 × 92 = 2³ × 3³ × 2² × 23¹
= 2⁴ × 3³ × 23¹

10. Simplify and write the following in the exponential form :

(i) 3³ × 2² + 2² × 5⁰
Solution:
3³ × 2² + 2² × 5⁰
= 3 × 3 × 3 × 2 × 2 + 2 × 2 × 5°
= 27 × 4 + 4 × 1
= 108 + 4
= 112
\(\begin{array}{c|c}
2 & 112 \\
\hline 2 & 56 \\
\hline 2 & 28 \\
\hline 2 & 14 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
= 2 × 2 × 2 × 2 × 7
= 2⁴ × 7¹

(ii) \(\left(\frac{3^{7}}{3^{2}}\right) \times 3^{5}\)
Solution:
\(\left(\frac{3^{7}}{3^{2}}\right) \times 3^{5}\) = (3^7-2) × 3⁵
= 3⁵ × 3⁵
= 3⁵⁺⁵
= 3¹⁰

(iii) 8² ÷ 2³
Solution:
8² ÷ 2³ = (2³)² ÷ 2³
= 2⁶ ÷ 2³
= 2^6-3
= 2³

Multiple Choice Questions :

11. \(\left(\frac{-5}{8}\right)^{0}\) is equal to :
(a) 0
(b) 1
(c) \(\frac {-5}{8}\)
(d) \(\frac {-8}{5}\)
Answer:
(b) 1

12. (5²)³ is equal to :
(a) 5⁶
(b) 5⁵
(c) 5⁹
(d) 10³
Answer:
(b) 5⁵

13. a × a × a × b × b × b is equal to :
(a) a³b²
(b) a²b³
(c) (ab)³
(d) a⁶b⁶
Answer:
(c) (ab)³

14. (-5)² × (-1)¹ is equal to :
(a) 25
(b) -25
(c) 10
(d) -10
Answer:
(b) -25