NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

Mathematics is a subject that requires students to have good reasoning and problem-solving skills. The only sure-shot way of getting good at Mathematics and scoring well in the final exams is to solve various problems of varying difficulty levels. In order to help students with this, Extramarks offers NCERT Solutions for Class 7 Mathematics Chapter 13. These solutions are prepared by subject matter experts that give special care and attention to providing accurate solutions.

Students can use these resources in order to prepare for their examinations, for last-minute revisions, and for help with their assignments.

NCERT Solutions for Class 7 Mathematics Chapter 13

Access NCERT Solutions for Class 7 Mathematics Chapter 13 – Exponents and Powers

Chapter 13 – Exponents and Powers Exercise
Exercise 13.1	Questions & Solutions
Exercise 13.2	Questions & Solutions
Exercise 13.3	Questions & Solutions

NCERT Solutions for Class 7 Mathematics Chapter 13

Chapter 13 of the Class 7 NCERT Mathematics textbook introduces students to exponents and powers. It starts by explaining how exponents are extremely useful for expressing large numbers easily. For example, 104 can be used to express 10,000 in a much more concise manner. Imagine expressing something like 1 followed by 20 zeroes without exponents.

In the given example 10 is the base and 4 is the exponent or the power.

The chapter then introduces students to the different laws of exponents. These laws are extremely important and are used extensively in Mathematics problems for simplifying complicated expressions. Students should give special attention to these laws and should solve as many problems as possible in order to get comfortable with them.

Class 7 Mathematics Chapter 13 Includes:

The different topics covered in Chapter 13 of Class 7 Mathematics NCERT textbook include:

Exponents
Laws of Exponents
Multiplying Powers With The Same Base
Dividing Powers With The Same Base
Taking Power of a Power
Multiplying Powers With The Same Exponents
Dividing Powers With The Same Exponents
Miscellaneous Examples Using The Laws of Exponents
Decimal Number System
Expressing Large Numbers in The Standard Form

Facts

-1 raise to an odd power will always be -1 and -1 raised to an even power is always 1
Let ‘a’ and ‘b’ be non-zero integers and m, n be whole numbers, then
- am x an = a(m+n)
- am ÷ an = a(m-n); m > n
- (am)n = am x n
- am x bm = (ab)m
- am ÷ bm = (a/b)m
- a0 = 1
Any decimal number between 1.0 and 10.0, including 1.0 multiplied by a power of ten, is said to be in standard form.
Exponents are used for writing long numbers as short notations to make these numbers easy to read, understand, and compare.

Exponents

Exponents simply represent a number getting multiplied by itself. Consider the following statements

x * x = x2

x * x * x = x3

x * x * x * x = x4

x * x * x * x * x * x = x6

The number xn is read as ‘x’ raised to the power of n or simply the nth power of ‘x ‘. Here, x is known as the base, and n is known as the exponent.

Laws of Exponents

Let us understand the laws of exponents.

Multiplying Powers With the Same Base

am x an = a(m+n)

where ‘a’ is any non-zero integer.

‘m’ and ‘n’ are whole numbers.

Example: 42 x 45 = 4(2+5) = 47

38 x 33 = 3(8+3) = 311

b2 x b5 = b(2+5) = b7

Dividing Powers with the Same Base

am ÷ an = a(m-n); m > n

where ‘a’ is any non-zero integer.

‘m’ and ‘n’ are whole numbers.

m > n

Example: 57 ÷ 53 = 5(7-3) = 54

210 ÷ 28 = 2(10-8) = 22

(-8)6 ÷ (-8)3 = (-8)6-3 = (-8)3

III. Taking Power of a Power

(am)n = amn

where ‘a’ is any non-zero integer.

‘m’ and ‘n’ are whole numbers.

Example: (23)5 = 23 x 5 = 215

(53)2 = 53 x 2 = 56

(-63)4 = (-6) 3 x 4 = (-6)12

Multiplying Powers with Same Exponents

am x bm = (ab)m

where ‘a’ is any non-zero integer.

‘m’ and ‘n’ are whole numbers.

Example: 23 x 33 = (2 x 3)3 = 63

52 x 62 = (5 x 6)2 = 302

(-2)4 x (-3)4 =(−2 x −3)4

Dividing the Powers with the Same Exponents

am ÷ bm = (a/b)m

where ‘a’ is any non-zero integer.

‘m’ and ‘n’ are whole numbers.

Example: 54 ÷ 34 =(5/3)4

(-3)2 ÷ p2 = (-3/p)2

Numbers with Exponent Zero

a0 = 1 ( for any non-zero integer ‘a’)

or

Any number (except 0) raised to the power ( or exponent) 0 is 1.

Example: (25)0 = 1

(½)0 = 1

(-18)0 = 1

NCERT Solutions for Class 7 Mathematics

Mathematics is one subject that requires a lot of practice and problem-solving skills to master and score good marks in the exams. With NCERT Solutions of Class 7 Mathematics, Extramarks tries to provide students with a valuable resource that they can use to get better at the subject and eventually score well in their final exams. Students can use these solutions as the answers are explained in detail, which gives students an idea of how to attempt a question in the exam in the right manner.

NCERT Solutions for Class 7

Apart from Mathematics, Extramarks provides detailed and authentic solutions to the textbook questions for all of the other subjects covered in Class 7. A team of subject matter experts has prepared these solutions while ensuring that it is highly accurate and easy to understand.

Students can download all these solutions from the Extramarks website or the app.

NCERT Solutions Class 7 Maths Chapter-wise List

Chapter 1 – Integers

Chapter 2 – Fractions and Decimals

Chapter 3 – Data Handling

Chapter 4 – Simple Equations

Chapter 5 – Lines and Angles

Chapter 6 – The Triangle and Its Properties

Chapter 7 – Congruence of Triangles

Chapter 8 – Comparing Quantities

Chapter 9 – Rational Numbers

Chapter 10 – Practical Geometry

Chapter 11 – Perimeter and Area

Chapter 12 – Algebraic Expressions

Chapter 13 – Exponents and Powers

Chapter 14 – Symmetry

Chapter 15 – Visualising Solid Shapes

Q.1 Show that 913×−1117−8=1398×17−118.

Ans.

$\begin{array}{l} {(\frac{9}{13} \times \frac{- 11}{17})}^{- 8} = {(\frac{13}{9})}^{8} \times {(\frac{17}{- 11})}^{8} \\ L . H . S = {(\frac{9}{13} \times \frac{- 11}{17})}^{- 8} \\ = {(\frac{13}{9})}^{- 8} \times {(\frac{- 11}{17})}^{- 8} \\ = {(\frac{13}{9})}^{8} \times {(\frac{17}{- 11})}^{8} \\ = R . H . S \end{array}$

Q.2 Simplify:

$(i) \frac{3^{2} \times 4^{5} \times x^{4}}{3^{4} \times 4^{3} \times x^{9}} (ii) \frac{4^{5} \times 9^{5} \times x^{7}}{2^{3} \times 3^{6} \times x^{5}}$

Ans.

$\begin{array}{l} (i) \frac{3^{2} \times 4^{5} \times x^{4}}{3^{4} \times 4^{3} \times x^{9}} \\ = 3^{(2 - 4)} \times 4^{(5 - 3)} \times x^{(4 - 9)} \\ = 3^{- 2} \times 4^{2} \times x^{- 5} \\ = \frac{4^{2}}{3^{2} \times x^{5}} \\ (ii) \frac{4^{5} \times 9^{5} \times x^{7}}{2^{3} \times 3^{6} \times x^{5}} \\ = \frac{{(2^{2})}^{5} \times {(3^{2})}^{5} \times x^{7}}{2^{3} \times 3^{6} \times x^{5}} \\ = \frac{2^{10} \times 3^{10} \times x^{7}}{2^{3} \times 3^{6} \times x^{5}} \\ = 2^{(10 - 3)} \times 3^{(10 - 6)} \times x^{(7 - 5)} \\ = 2^{7} \times 3^{4} \times x^{2} \end{array}$

Q.3 Express the following numbers in the expanded form.
(i) 5,223,000,000 (ii) 256,000,000

Ans.

$\begin{array}{l} (i) 5, 223, 000, 000 = 5 \times 10^{9} + 2 \times 10^{8} + 2 \times 10^{7} + 3 \times 10^{6} \\ (ii) 256, 000, 000 = 2 \times 10^{8} + 5 \times 10^{7} + 6 \times 10^{6} \end{array}$

Q.4 Find the number from each of the following expanded forms.

$\begin{array}{l} (i) 3 \times 10^{4} + 5 \times 10^{3} + 5 \times 10^{1} + 2 \times 10^{0} \\ (ii) 8 \times 10^{6} + 6 \times 10^{4} + 8 \times 10^{2} + 3 \times 10^{1} + 6 \times 10^{0} \end{array}$

Ans.

$\begin{array}{l} (i) 3 \times 10^{4} + 5 \times 10^{3} + 5 \times 10^{1} + 2 \times 10^{0} \\ = 30000 + 5000 + 50 + 2 \\ = 35052 \\ (ii) 8 \times 10^{6} + 6 \times 10^{4} + 8 \times 10^{2} + 3 \times 10^{1} + 6 \times 10^{0} \\ = 8000000 + 60000 + 800 + 6 \\ = 8060806 \end{array}$

Q.5 Express the following numbers in standard form.
(i) 296, 851, 358, 200 (ii) 25, 615, 646, 430

Ans.

$\begin{array}{l} (i) 296, 851, 358, 200 \\ = 2.96851358200 \times 10^{11} \\ (ii) 25, 615, 646, 430 \\ = 2.5 {615646430 × 10}^{10} \end{array}$

Q.6 Simplify and write the answer in exponential form.
(i) 3⁷÷ 3⁴ (ii) 5⁸ ÷ 5⁴

Ans.

$\begin{array}{l} (i) \frac{3^{7}}{3^{4}} = 3^{(7 - 4)} \\ = 3^{3} \\ (ii) \frac{5^{8}}{5^{4}} = 5^{(8 - 4)} \\ = 5^{4} \end{array}$

Q.7 Simplify and write the answer in exponential form.
(i) (6⁵)³÷ 6³ (ii) (9⁵⁰)³ (iii) (5³²)⁵ (iv) (2⁶⁴)⁵

Ans.

$\begin{array}{l} (i) \frac{{(6^{5})}^{3}}{6^{3}} = \frac{6^{15}}{6^{3}} \\ = 6^{(15 - 3)} \\ = 6^{12} \\ (ii) {(90^{50})}^{3} = {(90)}^{50 \times 3} \\ = {(90)}^{150} \\ (iii) {(5^{32})}^{5} = {(5)}^{32 \times 5} \\ = 5^{160} \\ (iv) {(2^{64})}^{5} = {(2)}^{64 \times 5} \\ = 2^{320} \end{array}$

Q.8 Find the value of (−9)³ × (−4)² .

Ans.

(−9)³ × (−4)²

=[(−9) × (−9) × (−9)] × [(−4) × (−4)]

= (−729) × (16)

= −11664

Q.9 Find the value of 60−20×60+20.

Ans.

$(6^{0} - 2^{0}) \times (6^{0} + 2^{0}) = (1 - 1) \times (1 - 1) = 0 \times 2 = 0$

Q.10 Express (3⁷x 3³) x 3³ as a rational number with negative exponent.

Ans.

$\begin{array}{l} (3^{7} \times 3^{3}) \times 3^{3} \\ = (3^{7 + 3}) \times 3^{3} \\ = 3^{10} \times 3^{3} \\ = {(3)}^{10 + 3} \\ = 3^{13} \\ = {(\frac{1}{3})}^{- 13} \end{array}$

Q.11 State true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³× 3² = 6⁵
(iii) 3⁰ = (1000)⁰

Ans.

(i) False;

10×10¹¹ = 10¹⁺¹¹= 10¹²

and, (100)¹¹ = (10²)¹¹= 10²²

(ii) False;

2³ = 8,

3² = 9

So, 2³ x 3² =72

and, 6⁵ = 6×6×6×6×6 = 7776

(iii) True;

3⁰ = 1 and (1000)⁰ = 1 .

Q.12

$Simplify \frac{2 \times 3^{4} \times 2^{5}}{9 \times 4^{2}} .$

Ans.

${\begin{array}{l} \frac{2 \times 3^{4} \times 2^{5}}{9 \times 4^{2}} \\ = \frac{2 \times 3^{4} \times 2^{5}}{3^{2} \times {(2^{2})}^{2}} \\ = \frac{2 \times 3^{4} \times 2^{5}}{3^{2} \times 2^{2 \times 2}} \\ = \frac{3^{4} \times 2^{1 + 5}}{3^{2} \times 2^{4}} \\ = \frac{3^{4} \times 2^{6}}{3^{2} \times 2^{4}} \\ = 2^{6 - 4} \times 3^{4 - 2} \\ = 2^{2} \times 3^{2} \\ = 4 \times 9 \\ = 36 \end{array}}^{}$

Q.13

$Express {(- 4)}^{- 1} \times {(\frac{1}{3})}^{- 1} as a rational number .$

Ans.

$\begin{array}{l} {(- 4)}^{- 1} \times {(\frac{1}{3})}^{- 1} = (\frac{1}{- 4}) \times \frac{1}{(1 / 3)} \\ = \frac{- 1}{4} \times 3 \\ = \frac{- 3}{4} \end{array}$

Q.14 Express the 16000 as a product of power of prime factors.

Ans.

16000

= 16 × 1000

= (2 ×2 × 2 × 2) × 1000

= 2⁴ × 10³ [as 16 = 2 ×2 × 2 × 2 ]

= (2 ×2 × 2 × 2) × (2 ×2 × 2 × 5 × 5 ×5) [as 10 = 2×5]

= (2 ×2 × 2 × 2 × 2 ×2 × 2 )× (5 × 5 ×5)

Hence, 16000 = 2⁷ × 5³

Q.15

$Simplify \frac{14^{4}}{7^{4.}}$

Ans.

$\begin{array}{l} \frac{14^{4}}{7^{4}} \\ = \frac{14 \times 14 \times 14 \times 14}{7 \times 7 \times 7 \times 7} \\ = \frac{14}{7} \times \frac{14}{7} \times \frac{14}{7} \times \frac{14}{7} \\ = {(\frac{14}{7})}^{4} \\ = {(2)}^{4} = 16 \end{array}$

Q.16 Simplify

$(2^{7} \times 2^{8}) \div 2^{12}$

Ans.

$\begin{array}{l} (2^{7} \times 2^{8}) \div 2^{12} = \frac{2^{7 + 8}}{2^{12}} \\ = \frac{2^{15}}{2^{12}} \\ = 2^{15 - 12} \\ = 2^{3} \end{array}$

Q.17 Simplify 2³×2²×5⁵ and write the answer in the exponential form.

Ans.

$\begin{array}{l} 2^{3} \times 2^{2} \times 5^{5} = 2^{5} \times 5^{5} \\ = {(2 \times 5)}^{5} \\ = 10^{5} \end{array}$

Q.18 Find the value of (4²)⁵.

Ans.

$\begin{array}{l} {(4^{2})}^{5} = 4^{2 \times 5} \\ = 4^{10} \end{array}$

Q.19 Simplify 7^x × 7² and write the answer in exponential form.

Ans.

7^x × 7² = 7^x+2

Q.20 Express 65,950 in the standard forms.

Ans.

65,950 = 6.595 × 10,000
= 6.595 × 10⁴

Q.21 Which one is greater 10² or 2¹⁰?

Ans.

10² = 10 × 10 = 100
2¹⁰ = (2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 ×2) = 1024
Since, 1024 > 100
Hence, 2¹⁰ > 10²

Q.22 Write 104278 in expanded form.

Ans.

104278 =1×100,000 + 0 × 10,000 + 4 × 1000 + 2 × 100 + 7 × 10 + 8 × 1

= 1×10⁵ + 0 × 10⁴ + 4 × 10³ + 2 × 10² + 7 × 10¹ + 8 × 10⁰

= 1×10⁵ + 4 × 10³ + 2 × 10² + 7 × 10¹ + 8 × 10⁰

Q.23 What number should be multiplied by (-8)^-1so that the product may be equal to (10)^-1?

Ans.

$\begin{array}{l} Let required number b x . \\ x \times {(- 8)}^{- 1} = 10^{- 1} \\ ∴ \\ \Rightarrow x = (10^{- 1}) \div {(- 8)}^{- 1} \\ x = (\frac{1}{10}) \div (\frac{- 1}{8}) \\ = \frac{1}{10} \times \frac{8}{- 1} \\ = \frac{- 4}{5} \end{array}$

Q.24 Expand a³b², a²b³, b²a³, b³a². Find out like terms.

Ans.

a³b² = a³× b²
= (a×a×a)×(b×b)
= a×a×a×b×b
a²b³ = a²× b³
=a×a×b×b×b
b²a³ = b²× a³
= b×b×a×a×a
b³a² = b³× a²
=b×b×b×a×a
b²a³ and a³b²are like terms, since the power of a and b in these two terms are the same. The order of factors do not matter.
Thus, a³b² = a³ × b² or b² × a³ = b²a³. Similarly, a²b³ and b³a² are like terms.

Q.25

$Simplify \frac{12^{4} \times 9^{3} \times 4}{6^{3} \times 8^{2} \times 27}$

Ans.

$\begin{array}{l} \frac{12^{4} \times 9^{3} \times 4}{6^{3} \times 8^{2} \times 27} \\ = \frac{{(2^{2} \times 3)}^{4} \times {(3^{2})}^{3} \times 2^{2}}{2^{3} \times 3^{3} \times {(2^{3})}^{3} \times 3^{3}} \\ = \frac{{(2^{2})}^{4} \times {(3)}^{4} \times 3^{2 \times 3} \times 2^{2}}{2^{3} \times 3^{3} \times 2^{3 \times 2} \times 3^{3}} \\ = \frac{2^{8} \times 2^{2} \times 3^{4} \times 3^{6}}{2^{3} \times 2^{6} \times 3^{3} \times 3^{3}} \\ = \frac{2^{8 + 2} \times 3^{4 + 6}}{2^{3 + 6} \times 3^{3 + 3}} \\ = \frac{2^{10} \times 3^{10}}{2^{9} \times 3^{6}} \\ = 2^{10 - 9} \times 3^{10 - 6} \\ = 2^{1} \times 3^{4} \\ = 2 \times 81 = 162 \end{array}$

Q.26

$Find m so that {(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{- 6} = {(\frac{2}{9})}^{2 m - 1}$

Ans.

$\begin{array}{l} {(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{- 6} = {(\frac{2}{9})}^{2 m - 1} \\ \Rightarrow {(\frac{2}{9})}^{3 - 6} = {(\frac{2}{9})}^{2 m - 1} \\ \Rightarrow {(\frac{2}{9})}^{- 3} = {(\frac{2}{9})}^{2 m - 1} \\ In an equation, when bases are same on both sides, powers must be equal . \\ ∴ 2 m - 1 = - 3 \\ 2 m = - 3 + 1 \\ 2 m = - 2 \\ m = \frac{- 2}{2} = - 1 \\ ∴ Required value of m = - 1 \end{array}$

Q.27

$If \frac{p}{q} = {(\frac{3}{2})}^{- 2} \div {(\frac{8}{27})}^{- 1 / 3}, find the value of {(\frac{p}{q})}^{- 3} .$

Ans.

$\begin{array}{l} If \frac{p}{q} = {(\frac{3}{2})}^{- 2} \div {(\frac{8}{27})}^{- 1 / 3} \\ \Rightarrow \frac{p}{q} = {(\frac{2}{3})}^{2} \div {(\frac{2}{3})}^{(- 1 / 3) x 3} \\ \Rightarrow \frac{p}{q} = {(\frac{2}{3})}^{2} \div {(\frac{2}{3})}^{- 1} \\ \Rightarrow \frac{p}{q} = \frac{4}{9} \times \frac{3}{2} \\ \Rightarrow \frac{p}{q} = (\frac{2}{3}) \\ \Rightarrow {(\frac{p}{q})}^{3} = {(\frac{2}{3})}^{3} = \frac{8}{27} \end{array}$

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FAQs (Frequently Asked Questions)

Yes, NCERT Solutions for Class 7 Mathematics Chapter 13 are available on the website and the app of Extramarks.

Yes, these solutions are reliable and are drafted by experienced faculty who have years of experience in their respective fields.

Exponents are essential because these deal with the study of huge numbers and their notations in short. Exponents have applications across all sorts of subjects ranging from physics, biology, astronomy, and many more.

NCERT Solutions for Class 7 Mathematics Chapter 13 Exponents and Powers

NCERT Solutions for Class 7 Mathematics Chapter 13

Access NCERT Solutions for Class 7 Mathematics Chapter 13 – Exponents and Powers

NCERT Solutions for Class 7 Mathematics Chapter 13

Class 7 Mathematics Chapter 13 Includes:

Facts

Exponents

Laws of Exponents

NCERT Solutions for Class 7 Mathematics

NCERT Solutions for Class 7

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NCERT Solutions for Class 7 Mathematics Chapter 13 Exponents and Powers

NCERT Solutions for Class 7 Mathematics Chapter 13

Access NCERT Solutions for Class 7 Mathematics Chapter 13 – Exponents and Powers

NCERT Solutions for Class 7 Mathematics Chapter 13

Class 7 Mathematics Chapter 13 Includes:

Facts

Exponents

Laws of Exponents

NCERT Solutions for Class 7 Mathematics

NCERT Solutions for Class 7

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Please register to view this section

FAQs (Frequently Asked Questions)

1. Are NCERT Solutions for Class 7 Mathematics Chapter 13 Exponents and Powers available online?

2. Are NCERT Solutions for Class 7 Mathematics Chapter 13 reliable?

3. Why are exponents important?

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