PSEB 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 13 Exponents and Powers Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

1. Fill in the blanks :

(i) In the expression 37, base = …………….. and exponent = ……………..
(ii) In the expression \(\left(\frac{2}{5}\right)^{11}\), base = …………….. and exponent = ……………..
Solution:
(i) 3, 7
(ii) \(\frac {2}{5}\), 11

2. Find the value of the following :
(i) 26
(ii) 93
(iii) 55
(iv) (-6)4
(v) \(\left(-\frac{2}{3}\right)^{5}\)
Solution:
(i) 26 = 2 × 2 × 2 × 2 × 2 × 2
= 64

(ii) 93 = 9 × 9 × 9
= 729

(iii) 55 = 5 × 5 ×5 × 5 × 5
= 3125

(iv) (-6)4 = -6 × -6 × -6 × -6
= 1296

(v) \(\left(-\frac{2}{3}\right)^{5}\) = \(\frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3}\)
= \(-\frac{32}{243}\)

PSEB 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1

3. Express the following in the exponential form :
(i) 6 × 6 × 6 × 6
(ii) b × b × b × b
(iii) 5 × 5 × 7 × 7 × 7
Solution:
(i) 6 × 6 × 6 × 6 = 64
(ii) b × b × b × b = b4
(iii) 5 × 5 × 7 × 7 × 7 = 52 × 73

4. Simplify the following :

(i) 2 × 103
Solution:
2 × 103 = 2 × 10 × 10 × 10
= 2000

(ii) 52 × 32
Solution:
52 × 32 = 5 × 5 × 3 × 3
= 25 × 9
= 225

(iii) 32 × 104
Solution:
32 × 104 = 3 × 3 × 10000
= 90000

PSEB 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1

5. Simplify :
(i) (-3) × (-2)3
Solution:
(-3) × (-2)3 = -3 × -2 × -2 × -2
= -3 × -8
= 24

(ii) (-4)3 × 52
Solution:
(-4)3 × 52= -4 × -4 × -4 × 5 × 5
= 64 × 25
= -1600

(iii) (-1)99
Solution:
(-1)99 = -1
[(-1)odd number = -1]

(iv) (-3)2 × (-5)2
Solution:
(-3)2 × (-5)2 = -3 × -3 × – 5 × -5
= 9 × 25
= 225

(v) (-1)132
Solution:
(-1)132 = 1
[(-1)even number = +1]

PSEB 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1

6. Identify the greater number in each of the following :

(i) 43 or 34
Solution:
43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
81 > 64
∴ 34 > 43.

(ii) 53 or 32
Solution:
53 = 5 × 5 × 5 = 125
32 = 3 × 3 = 9
125 > 9
∴ 53 > 32.

(iii) 23 or 82
Solution:
23 = 2 × 2 × 2 = 8
82 = 8 × 8 = 64
64 > 8
∴ 82 > 23.

(iv) 45 or 54
Solution:
45 = 4 × 4 × 4 × 4 × 4 = 1024
54 = 5 × 5 × 5 × 5 = 625
1024 > 625
∴ 45 > 54.

(v) 210 or 102
Solution:
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 1024
102 = 10 × 10 = 100
1024 > 100
∴ 210 > 102

PSEB 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1

7. Write the following numbers as power of 2 :

(i) 8
Solution:
8 = 2 × 2 × 2
\(\begin{array}{c|c}
2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
= 23

(ii) 128
Solution:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
= 27
\(\begin{array}{l|l}
2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)

(iii) 1024
Solution:
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
=210
\(\begin{array}{l|l}
2 & 1024 \\
\hline 2 & 512 \\
\hline 2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)

PSEB 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1

8. Write the following numbers as power of 3 :

(i) 27
Solution:
27 = 3 × 3 × 3
= 33
\(\begin{array}{l|l}
3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

(ii) 2187
Solution:
2187 = 3 × 3 × 3 × 3 × 3 × 3 × 3
= 37
\(\begin{array}{l|l}
3 & 2187 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

9. Find the value of x in each of the following:

(i) 7x = 343
Solution:
343 =7 × 7 × 7 = 73
7x = 343
7x = 73
∴ x = 3

(ii) 9x = 729
Solution:
729 =9 × 9 × 9
= 93
9x = 729
9x = 93
∴ x = 3.

(iii) (-8)x = -512
Solution:
512 = 8 × 8 × 8
= 83
(-8)x = -512
(-8)x = (-8)3
∴ x = 3.

PSEB 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1

10. To what power (-2) should be raised to get 16 ?
Solution:
Let power raised be x
16 = 2 × 2 × 2 × 2
= 24
(-2)x = 24
(-2)x = (-2)4
[(-1)even number = +1]
∴ x = 4.

11. Write the prime factorization of the following numbers in the exponential form :

(i) 72
Solution:
72 = 2 × 2 × 2 × 3 × 3
= 23 × 32
\(\begin{array}{l|l}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

(ii) 360
Solution:
360 = 2 × 2 × 2 × 3 × 3 × 5
= 23 × 32 × 51
\(\begin{array}{c|c}
2 & 360 \\
\hline 2 & 180 \\
\hline 2 & 90 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)

(iii) 405
Solution:
405 = 3 × 3 × 3 × 3 × 5
= 34 × 51
\(\begin{array}{l|l}
3 & 405 \\
\hline 3 & 135 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)

(iv) 648
Solution:
648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
= 23 × 34
\(\begin{array}{c|c}
2 & 648 \\
\hline 2 & 324 \\
\hline 2 & 162 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

(v) 3600
Solution:
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 24 × 32 × 52
\(\begin{array}{c|c}
2 & 3600 \\
\hline 2 & 1800 \\
\hline 2 & 900 \\
\hline 2 & 450 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)

Leave a Comment