Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 13 Exponents and Powers InText Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 13 Exponents and Powers InText Questions

Try These (Page 250)

Question 1.

Find five more such examples, where a number is expressed in exponential form. Also identify the base and the exponent in each case.

Solution:

Note:

1. x Ć x Ć x Ć x = x^{4} is read as āx raised to the power 4ā or ā4^{th} power of xā.

2. xĀ²y^{5} is read as āx squared into y raised to power 5ā.

3. p^{6}q^{3} is read as āp raised to the power 6 into q cubedā.

Try These (Page 251)

Question 1.

Express:

(i) 729 as a power of 3

(ii) 128 as a power of 2

(iii) 343 as a power of 7

Solution:

(i) 729

We have:

729 = 3 x 3 x 3 x 3 x 3 x 3 = 36

Thus, 729 = 3^{6}

(ii) 128

We have:

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^{7}

Thus, 128 = 2^{7}

(iii) 343

We have:

343 = 7 x 7 x 7 = 7^{3}

Thus 343 = 7^{3}

Try These (Page 254)

Question 1.

Simplify and write in exponential form:

(i) 2^{5} x 2^{3}

(ii) p^{3} x p^{2}

(iii) 4Ā³ x 4Ā²

(iv) a^{3} x a^{2} x a^{7}

(v) 5^{3} x 5^{7} x 5^{12}

(vi) (-4)^{100} x (4)^{20}

Solution:

Since a^{m} x a^{n} = a^{m+n}, therefore:

(i) 2^{5} x 2^{3}:

We have: 2^{5} x 2^{3} = 2^{5+3} = 2^{8}

(ii) p^{3} x p^{2}:

We have: p^{3} x p^{2} = p^{3+2} = p^{5}

(iii) 4^{3} x 4Ā²:

We have: 4Ā³ x 4Ā² = 4^{3+2} = 4^{5}

(iv) a^{3} x a^{2} x a^{7}

We have: a^{3} x a^{2} x a^{7} = a^{3+2+7} = a^{12}

(v) 5^{3} x 5^{7} x 5^{12}

We have: 5^{3} x 5^{7} x 5^{12} = 5^{3+7+12} = 5^{22}

(vi) (-4)^{100} x (-4)^{20}

We have: (-4)^{100} x (-4)^{20} = (-4)^{100+20} = (4)^{120}

Note: The above rule is possible only for same bases. It is not true for different bases. Thus, 2^{3} x 3Ā² will not obey this rule.

Try These (Page 255)

Question 1.

Simplify and write in exponential form: (example 11^{6} – 11^{2} = 11^{4} )

(i) 2^{9}Ć· 2^{3}

(ii) 10^{8} Ć· 10^{4}

(iii) 9^{11} Ć· 9^{7}

(iv) 20^{15} Ć· 20^{13}

(v) 7^{13} Ć· 7^{10}

Solution:

Since a^{m} Ć· a^{n} = a^{m-n}, therefore;

(i) 2^{9}Ć· 2^{3}:

We have: 2^{9} Ć· 2^{3} = 2^{9-3} = 2^{6}

(ii) 10^{8} Ć· 10^{4}:

We have: 10^{8} Ć· 10^{4} = 10^{8-4} = 10^{4}

(iii) 9^{11} Ć· 9^{7}

We have: 9^{11} Ć· 9^{7} = 9^{11-7} = 9^{4}

(iv) 20^{15} Ć· 20^{13}:

We have: 20^{15} Ć· 20^{13} = 20^{15-13} = 20^{2}

(v) 7^{13} Ć· 7^{10}:

We have: 7^{13} Ć· 7^{10} = 7^{13-10} = 7^{3}

Try These (Page 255)

Question 1.

Simplify and write the answer in exponential form:

(i) (6^{2})^{4}

(ii) (2Ā²)^{100}

(iii) (7^{50})Ā²

(iv) (5^{3})^{7}

Solution:

Since (a^{m})^{n} = a^{mĆn} = amn, therefore;

(i) (6^{2})^{4}:

We have: (6^{2})^{4} = 6^{2Ć4} = 6^{8}

(ii) (2Ā²)^{100}:

We have: (2Ā²)^{100} = 2Ā² x 100 = 2^{200}

(iii) 7^{50})Ā²:

We have: (7^{50})Ā² = 7^{50Ć2} = 7^{100}

(iv) (5^{3})7:

We have: (5^{3})7 = 5^{3Ć7} = 5^{21}

Try These (Page 256)

Question 1.

Put into another form using a^{m} x b^{m}Ā = (ab)^{m}:

(i) 4^{3} x 2^{3}

(ii) 2^{5} x b^{5}

(iii) a^{2} x t^{2}

(iv) 5^{6} x (-2)^{6}

(v) (-2)^{4} x (-3)^{4}

Solution:

(i) 4^{3} x 2^{3}:

We have: 4^{3} x 2^{3} = (4 x 2)^{3} = 8^{3}

(ii) 2^{5} x b^{5}:

We have: 2^{5} x b^{5} = (2 x b)^{5} = (2b)^{5}

(iii) a^{2} x t^{2}:

We have: a^{2} x t^{2} = (a x t)^{2} = (at)^{2}

(iv) 5^{6} x (-2)^{6}:

We have: 5^{6} x (-2)^{6} = [5 x (-2)]^{6} = (-10)^{6}

(v) (-2)^{4} x (-3)^{4}:

We have: (-2)^{4} x (-3)^{4} = [(-2) x (-3)]^{4} = [6]^{4}

Try These (Page 257)

Question 1.

Put into another form using a^{m} Ć· b^{n} = ( \(\frac { a }{ b }\) )^{m}.

(i) 4^{5} Ć· 3^{5}

(ii) 2^{5} Ć· b^{5}

(iii) (-2)^{3} Ć· b^{3}

(iv) p^{4} Ć· q^{4}

(v) 5^{6} Ć· (-2)^{6}

Solution:

Try These (Page 261)

Question 1.

Expand by expressing powers of 10 in the exponential form:

(i) 172

(ii) 5,643

(iii) 56,439

(iv) 1,76,428

Solution:

(i) 172:

We have:

172 = (1 x 100) + (7 x 10) + (2 x 1)

= 1 x 10Ā² + 7 x 10^{1} + 2 x 10^{0} (āµ 10Ā° = 1)

(ii) 5,643:

We have:

5,643 = 5 x 1000 + 6 x 100 + 4 x 10 + 3 x 1

= 5 x 10^{3} + 6 x 10^{2} + 4 x 10^{1} + 3 x 10^{0}

(iii) 56,439:

We have:

56,439 = 5 x 10000 + 6 x 1000 + 4 x 100 + 3 x 10 + 9 x 1

= 5 x 10^{4} + 6 x 10^{3} + 4 x 10^{2} + 3 x 10^{1} + 9 x 10^{0}

(iv) 1,76,428:

We have:

1,76,428 = 1,00,000 + 7 x 10,000 + 6 x 1000 + 4 x 100 + 2 x 10 + 8 x 1

= 1 x 10^{5} + 7 x 10^{4} + 6 x 10^{3} + 4 x 10^{2} + 2 x 10^{1} + 8 x 10^{0}