## NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers (EX 13.1) Exercise 13.1

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## Access NCERT Solution for Class 7 Chapter 13- Exponents and powers

Exercise 13.1

Refer to exercise 13.1 in the PDF

1. Find the value of:

i) \[{2^6}\]

Ans: The given expression is : \[{2^6}\]

By expanding the term \[{2^6}\] we get,

${2^6} = 2 \times 2 \times 2 \times 2 \times 2 \times 2$

By multiplying,

${2^6} = 64$

Hence, the required value of \[{2^6}\] is $64$.

ii) ${9^3}$

Ans: The given expression is : ${9^3}$

By expanding the term ${9^3}$ we get,

${9^3} = 9 \times 9 \times 9$

By multiplying,

${9^3} = 729$

Hence, the required value of ${9^3}$ is $729$.

iii) ${11^2}$

Ans: The given expression is : ${11^2}$

By expanding the term ${11^2}$ we get,

${11^2} = 11 \times 11$

By multiplying,

${11^2} = 121$

Hence, the required value of ${11^2}$ is $121$.

iv) ${5^4}$

Ans: The given expression is : ${5^4}$

By expanding the term ${5^4}$ we get,

${5^4} = 5 \times 5 \times 5 \times 5$

By multiplying,

${5^4} = 625$

Hence, the required value of ${5^4}$ is $625$.

2. Express the following in exponential form:

i) $6 \times 6 \times 6 \times 6$

Ans: The given expression is : $6 \times 6 \times 6 \times 6$

The number of $6$’s in the given expression is $4$ .

So, the exponent of $6$ is $4$.

Hence, $6 \times 6 \times 6 \times 6 = {6^4}$.

ii) $t \times t$

Ans: The given expression is : $t \times t$

The number of $t$ ’s in the given expression is $2$.

So, the exponent of $t$ is $2$.

Hence, $t \times t = {t^2}$.

iii) $b \times b \times b \times b$

Ans: The given expression is : $b \times b \times b \times b$

The number of $b$ ’s in the given expression is $4$.

So, the exponent of $t$ is $4$.

Hence, $b \times b \times b \times b = {b^4}$.

iv) $5 \times 5 \times 7 \times 7 \times 7$

Ans: The given expression is : $5 \times 5 \times 7 \times 7 \times 7$

The number of $5$ ’s in the given expression is $2$ .

The number of $7$ ’s in the given expression is $3$.

So, the exponent of $5$ is $2$ and the exponent of $7$ is $3$.

Hence, $5 \times 5 \times 7 \times 7 \times 7 = {5^2} \times {7^3}$.

v) $2 \times 2 \times a \times a$

Ans: The given expression is : $2 \times 2 \times a \times a$

The number of $2$ ’s in the given expression is $2$ .

The number of $a$ ’s in the given expression is $2$.

So, the exponent of $2$ is $2$ and the exponent of $a$ is $2$.

Hence, $2 \times 2 \times a \times a = {2^2} \times {a^2}$.

vi) $a \times a \times a \times c \times c \times c \times c \times d$

Ans: The given expression is : $a \times a \times a \times c \times c \times c \times c \times d$

The number of $a$ ’s in the given expression is $3$.

The number of $c$ ’s in the given expression is $4$.

The number of $d$ ’s in the given expression is $1$.

So, the exponent of $a$ is $3$ , the exponent of $c$ is $4$ and the exponent of $d$ is $1$ .

Hence, $a \times a \times a \times c \times c \times c \times c \times d = {a^3} \times {c^4} \times {d^1}$.

3. Express each of the following numbers using the exponential notation:

i) $512$

Ans: By factorization method, we can find,

(Image Will Be Updated Soon)

In the above expanding form of $512$, numbers of $2$’s as a factor is $9$.

So, we can express $512$ as term of $2$ .

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

The exponent of $2$ will be $9$ .

Hence, $512 = {2^9}$.

ii) $343$

Ans: By factorization method we can find,

7 | 343 |

7 | 49 |

7 | 7 |

1 |

In the above expanding form of $343$ , numbers of $7$’s as factor is $3$.

So, we can express $343$ as term of $7$ .

$343 = 7 \times 7 \times 7$

The exponent of $7$ will be $3$ .

Hence, $343 = {7^3}$.

iii) $729$

Ans: By factorization method we can find,

(Image Will Be Updated Soon)

In the above expanding form of $729$, numbers of $3$’s as the factor is $6$.

So, we can express $729$ as term of $3$ .

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$

The exponent of $3$ will be $6$ .

Hence, $729 = {3^6}$.

iv) $3125$

Ans: By factorization method, we can find,

(Image Will Be Updated Soon)

In the above expanding form of $3125$, numbers of $5$’s as a factor is $5$.

So, we can express $3125$ as term of $5$ .

$3125 = 5 \times 5 \times 5 \times 5 \times 5$

The exponent of $5$ will be $5$ .

Hence, $3125 = {5^5}$.

4. Identify the greater number, wherever possible, in each of the following:

i) ${4^3}$ and ${3^4}$

Ans: ${4^3}$ can be written as,

${4^3} = 4 \times 4 \times 4$

${4^3} = 64$

${3^4}$can be written as,

${3^4} = 3 \times 3 \times 3 \times 3$

${3^4} = 81$

Since, $81 > 64$

Thus, ${3^4} > {4^3}$.

Hence, ${3^4}$ is greater than ${4^3}$.

ii) ${5^3}$ and ${3^5}$

Ans: ${5^3}$ can be written as,

${5^3} = 5 \times 5 \times 5$

${5^3} = 125$

${3^5}$ can be written as,

${3^5} = 3 \times 3 \times 3 \times 3 \times 3$

${3^5} = 243$

Since, $243 > 125$

Thus, ${3^5} > {5^3}$.

Hence, ${3^5}$ is greater than ${5^3}$.

iii) ${2^8}$ and ${8^2}$

Ans: ${2^8}$ can be written as,

${2^8} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

${2^8} = 256$

${8^2}$ can be written as,

${8^2} = 8 \times 8$

${8^2} = 64$

Since, $256 > 64$

Thus, ${2^8} > {8^2}$.

Hence, ${2^8}$ is greater than ${8^2}$.

iv) ${100^2}$ and ${2^{100}}$

Ans: ${100^2}$ can be written as,

${100^2} = 100 \times 100$

${100^2} = 10000$

${2^{100}}$can be written as,

${2^{100}} = 2 \times 2 \times 2 \times 2 \times 2 \times ....14{\text{ times}} \times .... \times 2$

${2^{100}} = 16384 \times ... \times 2$

Since, $16384 \times ... \times 2 > 10000$

Thus, ${2^{100}} > {100^2}$.

Hence, ${2^{100}}$ is greater than ${100^2}$.

v) ${2^{10}}$ and ${10^2}$

Ans: ${2^{10}}$can be written as,

${2^{10}} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

${2^{10}} = 1024$

${10^2}$ can be written as,

${10^2} = 10 \times 10$

${10^2} = 100$

Since, $1024 > 100$

Thus, ${2^{10}} > {10^2}$.

Hence, ${2^{10}}$ is greater than ${10^2}$.

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

i) $648$

Ans: The given expression is $648$.

By factorization method we can find the prime factors,

(Image Will Be Updated Soon)

So, $648$ can be written as

$648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

It can be express as exponent of $2$ and $3$ .

$648 = {2^3} \times {3^4}$

Hence, the required expression is $648 = {2^3} \times {3^4}$.

ii) $405$

Ans: The given expression is $405$.

By factorization method we can find the prime factors,

(Image Will Be Updated Soon)

So, $405$ can be written as

$405 = 5 \times 3 \times 3 \times 3 \times 3$

Now, it can be express as

$405 = 5 \times {3^4}$

Hence, the required expression is $405 = 5 \times {3^4}$.

iii) $540$

Ans: The given expression is $540$.

By factorization method we can find the prime factors,

(Image Will Be Updated Soon)

So, $540$ can be written as

$540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5$

Now, it can be express as

$540 = {2^2} \times {3^3} \times 5$

Hence, the required expression is $540 = {2^2} \times {3^3} \times 5$.

iv) $3600$

Ans: The given expression is $3600$.

By factorization method we can find the prime factors,

(Image Will Be Updated Soon)

2 | 3600 |

2 | 1800 |

2 | 900 |

2 | 450 |

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

1 |

So, $3600$ can be written as

$3600 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$

Now, it can be express as

$3600 = {2^4} \times {3^2} \times {5^2}$

Hence, the required expression is $3600 = {2^4} \times {3^2} \times {5^2}$.

6. Simplify:

i) $2 \times {10^3}$

Ans: The given expression is: $2 \times {10^3}$

By expanding the given expression,

$2 \times {10^3} = 2 \times 10 \times 10 \times 10$

$2 \times {10^3} = 2000$

Hence, the required solution is $2 \times {10^3} = 2000$.

ii) ${7^2} \times {2^2}$

Ans: The given expression is: ${7^2} \times {2^2}$

By expanding the given expression,

${7^2} \times {2^2} = 7 \times 7 \times 2 \times 2$

${7^2} \times {2^2} = 196$

Hence, the required solution is ${7^2} \times {2^2} = 196$.

iii) ${2^3} \times 5$

Ans: The given expression is: ${2^3} \times 5$

By expanding the given expression,

${2^3} \times 5 = 2 \times 2 \times 2 \times 5$

${2^3} \times 5 = 40$

Hence, the required solution is ${2^3} \times 5 = 40$.

iv) $3 \times {4^4}$

Ans: The given expression is: $3 \times {4^4}$

By expanding the given expression,

$3 \times {4^4} = 3 \times 4 \times 4 \times 4 \times 4$

$3 \times {4^4} = 768$

Hence, the required solution is $3 \times {4^4} = 768$.

v) $0 \times {10^2}$

Ans: The given expression is: $0 \times {10^2}$

By expanding the given expression,

$0 \times {10^2} = 0 \times 10 \times 10$

$0 \times {10^2} = 0$

Hence, the required solution is $0 \times {10^2} = 0$.

vi) ${5^3} \times {3^3}$

Ans: The given expression is: ${5^3} \times {3^3}$

By expanding the given expression,

${5^3} \times {3^3} = 5 \times 5 \times 5 \times 3 \times 3 \times 3$

${5^3} \times {3^3} = 675$

Hence, the required solution is ${5^3} \times {3^3} = 675$.

vii) ${2^4} \times {3^2}$

Ans: The given expression is: ${2^4} \times {3^2}$

By expanding the given expression,

${2^4} \times {3^2} = 2 \times 2 \times 2 \times 2 \times 3 \times 3$

${2^4} \times {3^2} = 144$

Hence, the required solution is ${2^4} \times {3^2} = 144$.

viii) ${3^2} \times {10^4}$

Ans: The given expression is: ${3^2} \times {10^4}$

By expanding the given expression,

${3^2} \times {10^4} = 3 \times 3 \times 10 \times 10 \times 10 \times 10$

${3^2} \times {10^4} = 90,000$

Hence, the required solution is ${3^2} \times {10^4} = 90,000$.

7. Simplify:

i) ${\left( { - 4} \right)^3}$

Ans: The given expression is: ${\left( { - 4} \right)^3}$

By expanding the given expression,

${\left( { - 4} \right)^3} = \left( { - 4} \right) \times \left( { - 4} \right) \times \left( { - 4} \right)$

${\left( { - 4} \right)^3} = - 64$

Hence, the required solution is${\left( { - 4} \right)^3} = - 64$.

ii) $\left( { - 3} \right) \times {\left( { - 2} \right)^3}$

Ans: The given expression is: $\left( { - 3} \right) \times {\left( { - 2} \right)^3}$

By expanding the given expression,

$\left( { - 3} \right) \times {\left( { - 2} \right)^3} = \left( { - 3} \right) \times \left( { - 2} \right) \times \left( { - 2} \right) \times \left( { - 2} \right)$

$\left( { - 3} \right) \times {\left( { - 2} \right)^3} = 24$

Hence, the required solution is$\left( { - 3} \right) \times {\left( { - 2} \right)^3} = 24$.

iii) ${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2}$

Ans: The given expression is: ${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2}$

By expanding the given expression,

${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2} = \left( { - 3} \right) \times \left( { - 3} \right) \times \left( { - 5} \right) \times \left( { - 5} \right)$

${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2} = 225$

Hence, the required solution is${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2} = 225$.

iv) ${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3}$

Ans: The given expression is: ${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3}$

By expanding the given expression,

${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3} = 2 \times 2 \times 2 \times 10 \times 10 \times 10$

${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3} = 8000$

Hence, the required solution is${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3} = 8000$.

8. Compare the following numbers:

i) $2.7 \times {10^{12}};1.5 \times {10^8}$

Ans: The given expressions are :$2.7 \times {10^{12}}$and $1.5 \times {10^8}$.

Since, ${10^{12}} > {10^8}$

On comparing the exponents of base 10,

$2.7 \times {10^{12}} > 1.5 \times {10^8}$

Hence, $2.7 \times {10^{12}}$ is greater than $1.5 \times {10^8}$.

ii) $4 \times {10^{14}};3 \times {10^{17}}$

Ans: The given expressions are :$4 \times {10^{14}}$and $3 \times {10^{17}}$.

Since, ${10^{17}} > {10^{14}}$

On comparing the exponents of base 10,

$3 \times {10^{17}} > 4 \times {10^{14}}$

Hence, $3 \times {10^{17}}$ is greater than $4 \times {10^{14}}$.

## NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.1

Opting for the NCERT solutions for Ex 13.1 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.1 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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