# NCERT Solutions for Class 7 Mathematics 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

### 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.

1. 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

1. 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

1. 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

1. 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

1. 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.

Q.1 Show that 913×11178=1398×17118.

Ans.

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

Q.2 Simplify:

$\left(\mathrm{i}\right)\mathrm{ }\frac{{3}^{2}×{4}^{5}×{\mathrm{x}}^{4}}{{3}^{4}×{4}^{3}×{\mathrm{x}}^{9}} \left(\mathrm{ii}\right)\mathrm{ }\frac{{4}^{5}×{9}^{5}×{\mathrm{x}}^{7}}{{2}^{3}×{3}^{6}×{\mathrm{x}}^{5}}$

Ans.

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

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

Ans.

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

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

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }3×{10}^{4}+5×{10}^{3}+5×{10}^{1}+2×{10}^{0}\\ \left(\mathrm{ii}\right)\text{ }8×{10}^{6}+6×{10}^{4}+8×{10}^{2}+3×{10}^{1}+6×{10}^{0}\end{array}$

Ans.

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

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

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}}296,\text{ }851,\text{ }358,\text{ }200\\ =2.96851358200×{10}^{11}\\ \left(\mathrm{ii}\right)\text{ }25,\text{ 615, 646, 430}\\ \text{=}2.5{\text{615646430 × 10}}^{10}\end{array}$

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

Ans.

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

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

Ans.

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

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 6020×60+20.

Ans.

$\left({6}^{0}-{2}^{0}\right)×\left({6}^{0}+{2}^{0}\right)\phantom{\rule{0ex}{0ex}}=\left(1-1\right)×\left(1-1\right)\phantom{\rule{0ex}{0ex}}=0×2\phantom{\rule{0ex}{0ex}}=0$

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

Ans.

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

(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

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

Ans.

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

Q.13

$\mathrm{Express} {\left(-4\right)}^{-1}×{\left(\frac{1}{3}\right)}^{-1}\mathrm{as} \mathrm{a}\mathrm{rational}\mathrm{number}.$

Ans.

$\begin{array}{l}{\left(-4\right)}^{-1}×{\left(\frac{1}{3}\right)}^{-1}=\left(\frac{1}{-4}\right)×\frac{1}{\left(1/3\right)}\\ \text{ }=\text{ }\frac{-1}{4}\text{ }×3\\ \text{ }=\frac{-3}{4}\end{array}$

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

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

Ans.

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

Q.16 Simplify

$\left({2}^{7}×{2}^{8}\right)÷{2}^{12}$

Ans.

$\begin{array}{l}\left({2}^{7}×{2}^{8}\right)÷{2}^{12}=\frac{{2}^{7+8}}{{2}^{12}}\\ \text{ }=\frac{{2}^{15}}{{2}^{12}}\\ \text{ }={2}^{15-12}\\ \text{ }={2}^{3}\end{array}$

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

Ans.

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

Q.18 Find the value of (42)5.

Ans.

$\begin{array}{l}{\left({4}^{2}\right)}^{5}={4}^{2×5}\\ \mathrm{ }={4}^{10}\end{array}$

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

Ans.

7x × 72 = 7x+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)-1 so that the product may be equal to (10)-1?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{required}\mathrm{number}\mathrm{b}\mathrm{x}.\\ \mathrm{x}×{\left(-8\right)}^{-1}={10}^{-1}\\ \therefore \\ ⇒\mathrm{ }\mathrm{x}=\left({10}^{-1}\right)÷{\left(-8\right)}^{-1}\\ \mathrm{x}=\left(\frac{1}{10}\right)÷\left(\frac{-1}{8}\right)\\ =\frac{1}{10}×\frac{8}{-1}\\ =\frac{-4}{5}\end{array}$

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 a3b2 are 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

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

Ans.

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

Q.26

$\text{Find m so that }{\left(\frac{2}{9}\right)}^{3}×{\left(\frac{2}{9}\right)}^{-6}={\left(\frac{2}{9}\right)}^{2m-1}$

Ans.

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

Q.27

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

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{ }\frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{3}{2}\right)}^{-2}÷{\left(\frac{8}{27}\right)}^{-1/3}\\ ⇒\mathrm{ }\frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{2}{3}\right)}^{2}÷{\left(\frac{2}{3}\right)}^{\left(-1/3\right)\mathrm{x}3}\\ ⇒\mathrm{ }\frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{2}{3}\right)}^{2}÷{\left(\frac{2}{3}\right)}^{-1}\\ ⇒\mathrm{ }\frac{\mathrm{p}}{\mathrm{q}}=\frac{4}{9}×\frac{3}{2}\\ ⇒\mathrm{ }\frac{\mathrm{p}}{\mathrm{q}}=\left(\frac{2}{3}\right)\\ ⇒\mathrm{ }{\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^{3}={\left(\frac{2}{3}\right)}^{3}=\frac{8}{27}\end{array}$