These NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions

NCERT In-text Question Page No. 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.

Answer:

Number | Exponential form | Base | Exponent |

(i) 243 = 3 × 3 × 3 × 3 × 3 | 3^{5} |
3 | 5 |

(ii) 625 = 5 × 5 × 5 × 5 | 5^{4} |
5 | 4 |

(iii) 343 = 7 × 7 × 7 . | 7^{3} |
7 | 3 |

(iv) 1331 = 11 × 11 × 11 | 11^{3} |
11 | 3 |

(v) 64 = 8 × 8 | 8^{2} |
8 | 2 |

Note: 1. xxxxxxx = x4is read as ‘x raised to the power 4’ or ‘4^{th} power of x’.

2. x^{2}y^{5} is read as ‘x squared into y raised to power 5’.

3. p^{6}q^{3} is reas as ‘p raised to the power 6 into q cubed’.

NCERT In-text Question Page No. 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

Answer:

(i) 729

We have: 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3^{6}

Thus, 729 = 3^{6}

(ii) 128

We have: 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^{7}

Thus, 128 = 2^{7}

(iii) 343

We have: 343 = 7 × 7 × 7 × 7= 7^{3}

Thus, 343 = 7^{3}

NCERT In-text Question Page No. 254

Question 1.

Simplify and write in exponential form:

(i) 2^{5} × 2^{3}

(ii) p^{3} × p^{2}

(iii) 4^{3} × 4^{2}

(iv) a^{3} × a^{2} × a^{7}

(v) 5^{3} × 5^{7} × 5^{12}

(vi) (-4)^{100} × (-4)^{20}

Answer:

(i) 2^{5} × 2^{3}

We have: 2^{5} × 2^{3} = 2^{5 + 3} = 28

(ii) p^{3} × p^{2}

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

(iii) 4^{3} × 4^{2}

We have: 4^{3} × 4^{2<.sup> = 43+2 = p5}

(iv) a^{3} × a^{2} × a^{7}

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

(v) 5^{3} × 5^{7} × 5^{12}

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

(vi) (-4)^{100} × (-4)^{20}

We have: (-4)^{100} × (-4)^{20} = (-4)^{100+2} = (4)^{120}

Note: The above rule is possible only for same bases. It is not true for different bases. Thus, 2^{3} x 3^{2} will no obev this rule.

NCERT In-text Question Page No. 255

Question 1.

Simplify and write in exponential form: (eg., 11^{6} ÷ 11^{2} = 11^{4})

(i) 2^{9} ÷ 2^{3}

(ii) 10^{8} ÷ 10^{4}

(iii) 9^{11} ÷ 9^{7}

(iv) 20^{15} ÷ 20^{13}

(v) 7^{13} ÷ 7^{10}

Answer:

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

(i) 2^{9} ÷ 2^{3}

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

(ii) 10^{8} ÷ 10^{4}

We have: 10^{8} ÷ 10^{4} = 10^{8-4} = 10^{4}

(iii) 9^{11} ÷ 9^{7}

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

(iv) 20^{15} ÷ 20^{13}

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

(v) 7^{13} ÷ 7^{10}

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

Question 15.

Simplify and write the answer in exponential form:

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

(ii) (2^{2})^{100}

(iii) (7^{50})^{2}

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

Answer:

Since (a^{m})^{n} = a^{mxn} = a^{mn}, therefore;

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

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

(ii) (2^{2})^{100}

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

(iii) (7^{50})^{2}

We have: (7^{50} )^{2} = 7^{50 × 2} = 7^{100}

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

We have: (5^{3})^{7} = 5^{3 × 7} = 5^{21}

NCERT In-text Question Page No. 256

Question 1.

Put into another form using a^{m} × b^{m}= (ab)^{m}

(i) 4^{3} × 2^{3}

(ii) 2^{5} × b^{5}

(iii) a^{2} × t^{2}

(iv) 5^{6} × (-2)^{6}

(v) (-2)^{4} × (-3)^{4}

Answer:

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

(i) 4^{3} × 2^{3}

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

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

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

(iii) a^{2} × t^{2}

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

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

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

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

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

NCERT In-text Question Page No. 257

Question 2.

Put into another form using a^{m} ÷ b^{m}

(i) 4^{5} ÷ 3^{5}

(ii) (-2)^{5} ÷ b^{5}

(iii) (-2)^{3} ÷ b^{3}

(iv) p^{4} ÷ q^{4}

(v) 5^{6} ÷ (-2)^{6}

Answer:

(i) 4^{5} ÷ 3^{5}

We have: 4^{5} ÷ 3^{5} = \(\left(\frac{4}{3}\right)^{5}\)

(ii) (-2)^{5} ÷ b^{5}

We have: (-2)^{5} ÷ b^{5} = \(\left(\frac{2}{b}\right)^{5}\)

(iii) (-2)^{3} ÷ b^{3}

We have: (-2)^{3} ÷ b^{3} = \(\left(\frac{-2}{b}\right)^{3}\)

(iv) p^{4} ÷ q^{4}

We have: p^{4} ÷ q^{4} = \(\left(\frac{p}{q}\right)^{4}\)

(v) 5^{6} ÷ (-2)^{6}

We have: 5^{6} ÷ (-2)^{6} = \(\left(\frac{5}{-2}\right)^{6}\) = \(\left(-\frac{5}{2}\right)^{6}\)

NCERT In-text Question Page No. 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

Answer:

(i) 172:

We have:

172 =(1 × 100) + ( 7 × 10) + (2 × 1)

= 1 × 10^{2} + 7 × 10^{1} + 2 × 1)

(∵10^{0} = 1)

(ii) 5,643:

We have:

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

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

(iii) 56,439:

We have:

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

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

(iv) 1,76,428:

We have:

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

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