Students can Download Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

Question 1.

Find the value of:

(i) 2^{6}

(ii) 9^{3}

(iii) 11^{2}

(iv) 5^{4}

Solution:

(i) 2^{6} = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9^{3} = 9 × 9 × 9 = 729

(iii) 11^{2}= 11 × 11 = 121

(iv) 5^{4} = 5 × 5 × 5 × 5 = 625

Question 2.

Express the following in exponent form:

(i) 6 × 6 × 6 × 6

(ii) t × t

(iii) b × b × b × b

(iv) 5 × 5 × 7 × 7 × 7

(v) 2 × 2 × a × a

(vi) a × a × a × c × c × c × c × d

Solution:

(i) 6 × 6 × 6 × 6 = 6^{4}

(ii) t × t = t^{2}

(iii) b × b × b × b = b^{4}

(iv) 5 × 5 × 7 × 7 × 7 = 5^{2} × 7^{3}

(v) 2 × 2 × a × a = 2^{2} × a^{2}

(vi) a × a × a × c × c × c × c × d = a^{3} × c^{4} × d

Question 3.

Express each of the following numbers using the exponential notation:

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Solution:

(i) 512

Prime factorizing

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2^{9}

∴ 512 = 2^{9}

(ii) 343

Prime factorizing

= 7 × 7 × 7

Hence, 343 = 7^{3}

(iii) 729

Prime factorizing

= 3 × 3 × 3 × 3 × 3 × 3

= 3^{6}

Hence, 729 = 3^{6}

(iv) 3125

Prime factorizing

= 5 × 5 × 5 × 5 × 5 × 5

= 5^{5}

Question 4.

Identify the greater number, wherever possible, in each of the following?

(i) 4^{3} or 3^{4}

(ii) 5^{3} or 3^{5}

(iii) 2^{8} or 8^{2}

(iv) 100^{2} or 2^{100}

(v) 2^{10} or 10^{2}

Solution:

(i) 4^{3} or 3^{4}

4^{3} = 4 × 4 × 4 = 64

3^{4} = 3 × 3 × 3 × 3 = 81

81 > 64

Hence 3^{4} > 4^{3}

(ii) 5^{3} or 3^{5}

5^{3} = 5 × 5 × 5 = 125

3^{5} = 3 × 3 × 3 × 3 × 3 = 243

243 > 125

Hence 3^{5} > 5^{3}

(iii) 2^{8} or 8^{2}

2^{8} = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

8^{2} = 8 × 8 = 64

256 > 64

Hence 2^{8} > 8^{2}

(iv) 100^{2} or 2^{100}

Clearly 2^{100}

(v) 2^{10} or 10^{2}

2^{10} = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

10^{2} = 10 × 10= 100

1024 > 100

Hence 2^{10} > 10^{2}

Question 5.

Express each of the following as product of powers of their prime factors:

(i) 648

(ii) 405

(iii) 540

(iv) 3,600

Solution:

(i) 648

Prime factorizing

2 × 2 × 2 × 3 × 3 × 3 × 3

648 = 2^{3} × 3^{4}

(ii) 405

Prime factorizing

3 × 3 × 3 × 3 × 5

Hence, 405 = 3^{4} × 5

(iii) 540

Prime factorizing

2 × 2 × 3 × 3 × 3 × 5

540 = 2^{2} × 3^{3} × 5

(iv) 3600

Prime factorizing

= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

3600 = 2^{4} × 3^{2} × 5^{2}

Question 6.

Simplify:

(i) 2 × 10^{3}

(ii) 7^{2} × 2^{2}

(iii) 2^{3} × 5

(iv) 3 × 4^{4}

(v) 0 × 10^{2}

(vi) 5^{2} × 3^{3}

(vii) 2^{4} × 3^{2}

(viii) 3^{2} × 10^{4}

Solution:

(i) 2 × 10^{3}

10^{3} = 10 × 10 × 10 = 1000

2 × 1000 = 2000

(ii) 7^{2} × 2^{2}

7^{2} = 7 × 7 = 49, 2^{2} = 2 × 2=4

49 × 4=196

(iii) 2^{3} × 5

2^{3} = 2 × 2 × 2=8

8 × 5 = 40

(iv) 3 × 4^{4}

4^{4} = 4 × 4 × 4 × 4 = 256

23 × 256 = 768

(v) 0 × 10^{2} = 0

(vi) 5^{2} × 3^{3}

5 × 5 × 3 × 3 × 3

25 × 27 = 675

(vi) 2^{4} × 3^{2}

2 × 2 × 2 × 2 × 3 × 3

16×9= 144

(vii) 32 × 104

3 × 3 × 10 × 10 × 10 × 10

9 × 10000 = 90000

Question 7.

Simplify:

(i) (- 4)^{3}

(ii) (- 3) × (- 2)^{3}

(iii) (- 3)^{2} × (- 5)^{2}

(iv) (- 2)^{3} × (- 10)^{3}

Solution:

(i) (- 4)^{3}

(- 4) × (- 4) × (- 4) = – 64

(ii) (- 3) × (- 2)^{3}

(- 2)^{3} = (- 2) × (- 2) × (- 2) = – 8

(- 3) × (-8) = 24

(iii) (- 3)^{2} × (- 5)^{2}

(-3)^{2} = (- 3) × (- 3) = 9

(- 5)^{2} = (- 5) × (- 5) = 25

9 × 25 = 225

(iv) (- 2)^{3} × (- 10)^{3}

(- 2)^{3} = (- 2) × (- 2) × (- 2) = – 8

(- 10)^{3} = (- 10)(- 10)(- 10) = – 1000

– 8 × (- 1000) = 8000

Question 8.

Compare the following numbers:

(i) 2.7 × 10^{12}, 1.5 × 10^{8}

(ii) 4 × ^{14} , 3 × 10^{17}

Solution:

(i) 2.7 × 10^{12} = 2.7 × 1000000000000

= 2700000000000

1.5 × 10^{8} = 1.5 × 100000000

= 150000000

2.7 × 10^{2} > 1.5 × 10^{8}

(ii) 4 × 10^{14}, 3 × 10^{17}

4 × 10^{14} = 400000000000000

3 × 10^{17} = 300000000000000000

Hence, 4 × 10^{14} < 3 × 10^{17}

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