Class 7 Maths Chapter 13 Exponents and Powers

Exercise-13.1

Question 1:

Find the value of:

(i) 2⁶

(ii) 9³

(iii) 11²

(iv) 5⁴

Answer:

(i)In the given question,

We have to find the value of 26

We have,

26 = 2 × 2 × 2 × 2 × 2 × 2= 64

(ii) In the given question,

We have to find the value of 93

We have,

93 = 9 × 9 × 9 = 729

(iii) In the given question,

We have to find the value of 112

We have,

112 = 11 × 11 = 121

(iv) In the given question,

We have to find the value of 54

We have,

54 = 5 × 5 × 5 × 5= 625

Question 2:

Express the following in exponential form:

(i) 6 × 6 × 6 × 6      

(ii) t × t

(iii) b × b × b × b   

(iv) 5 × 5 × 7 × 7 × 7

(v) 2 × 2 × a × a    

(vi) a × a × a × c × c × c × d

Answer:

(i) In the given question,

We have to express the given expression into exponential form

We have,

6 × 6 × 6 × 6 = 64

(ii) In the given question,

We have to express the given expression into exponential form

We have,

t × t = t2

(iii) In the given question,

We have to express the given expression into exponential form

We have,

b × b × b × b = b4

(iv) In the given question,

We have to express the given expression into exponential form

We have,

5 × 5 × 7 × 7 × 7 = 52 × 73

(v) In the given question,

We have to express the given expression into exponential form

We have,

2 × 2 × a × a = 22 × a2

(vi) In the given question,

We have to express the given expression into exponential form

We have,

a × a × a × c × c × c × c × d = a3 × c4 × d

Question 3:

Express each of the following numbers using the exponential notation:

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Answer:

(i) In the given question,

We have to express the given numbers into exponential notation

We have,

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29

(ii) In the given question,

We have to express the given numbers into exponential notation

We have,

343 = 7 × 7 × 7 = 73

(iii) In the given question,

We have to express the given numbers into exponential notation

We have,

729 = 3 × 3 × 3 × 3 × 3 × 3 = 36

(iv) In the given question,

We have to express the given numbers into exponential notation

We have,

3125 = 5 × 5 × 5 × 5 × 5 = 55

Question 4:

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

Exercise-13.2

Question 1:

Using laws of exponents simplify and write the answer in exponential form:

(i) 3² × 3⁴ × 3⁸

(ii) 

(iii)  a³  × a²

(iv) 7^x × 7²

(v)        

(vi)2⁵ × 5²

(vii)a⁴ × b⁴

(viii)(3⁴)³

(ix)

(x)

Answer:

(i) We have,

3²× 3⁴ × 3⁸

We know that,

(a^m× aⁿ = a^{m + n})

Thus,

3²× 3⁴ × 3⁸

= (3)^{2 + 4 + 8}

= 3¹⁴

(ii) We have,

6¹⁵ 6¹⁰

We know that,

(a^m aⁿ = a^{m – n})

Thus,

6¹⁵ 6¹⁰

= (6)^{15 – 10}

= 6⁵

(iii) We have,

a³× a²

We know that,

(a^m× aⁿ = a^{m + n})

Therefore,

a³× a²

= (a)^{3 + 2}

= a⁵

(iv) We have,

7^x × 7²

We know that,

(a^m × aⁿ = a^{m + n})

Thus,

7^x × 7²

= (7)^{x + 2}

(v) We have,

(5²)³ 5³

Using identity:

(a^m)ⁿ= a^m × n

= 5^{2 × 3} 5³

= 5⁶ 5³

We know that,

(a^m aⁿ = a^{m – n})

Thus,

5⁶ 5³

= (5)^{6 – 3}

= 5³

(vi) We have,

2⁵× 5⁵

We know that,

[a^m ×b^m = (a × b)^m]

Thus,

2⁵× 5⁵

= (2 × 5)^{5 + 5}

= 10⁵

(vii) We have,

a⁴ × b⁴

We know that,

[a^m × b^m = (a × b)^m]

Thus,

a⁴ × b⁴

= (a × b)⁴

(viii) We have,

(3⁴)³

We know that,

(a^m)n = a^mn)

Thus,

(3⁴)³

= (3⁴)³

= 3¹²

(ix) We have,

(2²⁰ 2¹⁵) × 2³

We know that,

(a^m aⁿ = a^{m – n})

Thus,

(2^{20 – 15}) × 2³

= (2)⁵ × 2³

We know that,

(a^m × aⁿ = a^{m + n})

Thus,

(2)⁵ × 2³

=(2^{5 + 3})

= 2⁸

(x) We have,

(8^t 8²)

We know that,

(a^m aⁿ = a^{m – n})

Thus,

(8^t 8²)

= (8^{t – 2})

Exercise-13.3