Frank ICSE Class 8 Solutions Chapter 2

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Frank ICSE Mathematics Class 8 Solutions Chapter 2 Exponents

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 2, Exponents. Here students can easily find step by step solutions of all the problems for Exponents, Exercise 2.1 and 2.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here in this post all the solutions are based on latest Syllabus.

Exponents Exercise 2.1 Solution :

Question no – (1) 

Solution :

(a) 6^-9

= 1/6⁹

(b) 4^-11

= 4^-11
= 1/4¹¹

(c) (3/4)^-5

= (4/3)5

(d) (-4/7)^-12

= (- 4/7)^-12

= (7/-4)²

(e) (9/10)^-13

= (9/10)^-13

= (10/9)¹³

Question no – (2)

Solution :

(a) 5^-7

= 5^-8

= 1/5⁸

(b) (54)

= 5¹² ÷ 5

= 5¹³

Question no – (3) 

Solution :

= 8³ is the multiplicative inverse of 8^-3

Question no – (4)

Solution :

= 10¹⁰⁰ ÷ 10 is the same as 10⁹⁹

Question no – (5)

Solution :

(a) 64 in exponential form

= 2⁶

(b) 1/243 in exponential form

= 3^-5

(c) -125/8 in exponential form

= (- 5/2)³

(d) 1296 in exponential form

= 6⁴

(e) 16/81 in exponential form

= (2/3)⁴

(f) 625/256 in exponential form

= (5/4)⁴

(g) – 81/16 in exponential form

= (- 3/2)⁴

(h) – 216/343 in exponential form

= (- 6/7)³

Question no – (6) 

Solution :

(a) (2)^-3

= 1/8

∴ The value is 1/8

(b) (- 3)⁴

= 81

∴ The value is 81

(c) (- 5)^-3

= 1/-125

∴ The value is 1/-125

(d) (1/2)⁵

= 1/32

∴ The value is = 1/32

(e) (1/3)^-3

= 27

∴ The value is 27

Question no – (7) 

Solution :

(a) We have 28 × 10¹⁹ and 3 × 10²⁰

During both with 1019 we get 28 and (3 × 10) = 30

Among then 28 <3a

So, clearly, 2 × 10²⁰ is larger than 8 × 10^-13

(b) We have 2 × 10^-12 and 8 × 10^-13

Multiplying both with 10¹³, we get 2 × 10 and 8

Among them 20 > 8

So, clearly 2 × 10^-12 is greater than 8 × 10^-13

Question no – (8) 

Solution :

(a) Given, (5)³ × (5)^-8 × (5)^-11

= (5)3-8-11

= (5)^-15….(Simplified)

(b) Given, (-6)⁴ × 3⁵ ÷ 2^-4

= (- 6)⁴ × 3⁵ × 24

= 3⁹ × 2⁸  ….(Simplified)

(c) Given, 3² × 2^-3 × (1/4)^-2

= 3² × 2^-3 × 2⁴

= 3² × 2

= 18 ….(Simplified)

(d) Given, (5)^-4 × (4/5)^-4 × 2⁸

= (4)^-4 × 2⁸

= 2^-8 × 2⁸

= 1 ….(Simplified)

(e) Given, (-3)⁴ × (- 3)⁸

= (- 3)⁹ × (- 3)^-4

= (-3)^9-4

= (- 3)⁵ ….(Simplified)

Question no – (9)

Solution :

(a) Given in the question,

= {(1/3)^-3 – (1/2)^-3} ÷ (1/4)^-3

Now,

= {(1/3)^-3 – (1/2)^-3} ÷ (1/4)^-3

= {33 – 23} × 43

= (27 – 8) × 64

= 19 × 64

= 1216

Thus, the value is 1216.

(b) In the question

= {(1/3)^-1 × (- 9)^-1}^-1

Now,

= {(1/3)^-1 × (- 9)^-1}^-1

= {3 × 1/9}^-1

= {1/-3}^-1

= – 3

Therefore, the value is -3

(c) From the question we get,

= (5² – 3²) × (2/3)²

Now,

= (5² – 3²) × (2/3)²

= (25 – 9) × 4/9

= 16 × 4/9

= 64/9

Hence, the value is 64/9

(d) In the question,

= {(2/3)²}³ × (1/3)^-4 × (3)^-2 × (2)^-1

Now,

= {(2/3)²}³ × (1/3)^-4 × (3)^-2 × (2)^-1

= 64/729 × 81 × 1/9 × 1/2

= 32/81

Therefore, the value is 32/81

(e) (4^-1 ÷ 6^-1)³

= (1/4 × 6)³

= (3/2)³

= 27/8

Therefore, the value is 27/8

Question no – (10)

Solution :

(a) (2^-1 + 3^-1 + 4^-1)⁰

= (1/2 + 1/3 + 1/4)⁰

= (13/12)⁰

= 1

(b) (2^-1 + 3^-1 + 4^-1)⁰

= (1/2 + 1/3 + 1/4)⁰

= (13/12)⁰

= 1

(c) In the given question,

= {(2° + 3^-1) × 9²}

Now,

= {(2⁰ + 3-1) × 9²}

= {1/3 × 9²}

= 27

(e) In the given question,

= (5/7)^-6 × (49/25)^-4

Now,

= (5/7)^-6 × (49/25)^-4

= 25/49

(f) From the question we get,

= (4/9)^-3 × (2/3)¹⁰ × (8/27)^-2

Now,

= (4/9)^-3 × (2/3)¹⁰ × (8/27)^-2

= (2/3)^-6+10-6

= (2/6)^-2

= 9/4

Question no – (11) 

Solution :

(a) Given, 16a³b^-4c3/24a^-2b²c^-4

= 2/3 a⁵b^-6c^-1

(b) Given, 25a^-2b²c²/6a4b^-4c^-2

= 5/12 a^-6b⁶c⁴

(c) Given, (a²ⁿ × a³ⁿ × a⁴ⁿ)⁵

= (a⁹ⁿ)⁵

= a⁴⁵ⁿ

Exponents Exercise 2.2 Solution :

Question no – (1) 

Solution :

(a) 2^x = 1/64

or, 2^x = 1/64

or, 2x = 1/2⁶

or, 2x = 2^-6

or, x = – 6

Therefore, x = -6

(b) 16^2x-1 = 8^x+2

Now, 16^2x-1 = 8^x+2

or, (2)8x-4 = (2)3x + 6

or, 8x – 4 = 3x + 6

or, 5x = 10

or, x = 2

Thus, x = 2

(c) 125^x-1 = 25^x+2

Now, 125^x-1 = 25^x+2

= or, (5)^3x-3 = (5)^2x+4

or, 3x – 3 = 2x + 4

or, 3x – 3 = 2x + 4

or, x = 7

Hence, x = 7

(d) 125^x-1 = 25^x+2

Now, 125^x-1 = 25^x+2

= or, (5)^3x-3 = (5)^2x+4

or, 3x – 3 = 2x + 4

or, 3x – 3 = 2x + 4

or, x = 7

Hence, x = 7

(e) 7^3x+1 = 1/49

Now, 7^3x+1 = 1/49

or, 7^3x+1 = 1/7²

or, 7^3x+1 = 7^-2

or, 3x + 1 = – 2

or, 3x = – 3

or, x = – 1

Therefore, x = – 1

(f) 6^2x-1 × 36^3x-2 = 216

Now, 6^2x-1 × 36^3x-2 = 216

or, (6)^2x-1 × (6)^6x-4 = 6³

or, (6)^2x-1+6x-4 = 6³

or, 8x – 5 = 3

or, 8x = 8

or, x = 1

Hence, x = 1

(g) (3/4)^x+2 = (27/64)^4-3x

Now, (3/4)^x+2 = (27/64)^4-3x

= (3/4)^x+2 = (3/4)^12-9x

or, x + 2 = 12 – 9x

or, 10x = 10

or, x = 1

Thus, x = 1

(h) (3^-x)^-2 × 3^x × 81 = 1/9

Now, = (3^-x)^-2 × 3^x × 81 = 1/9

or, (3)^2x × 3^x × (3)⁴ = 1/3²

or, 3^2x+x+4 = 3^-2

or, 3x + 4 = – 2

or, 3x = 6

or, x = 2

Therefore, x = 2

(i) 7^1–2x × 49³ × 343 = 1/7^-4

or, 7^1-2x × (7)⁶ × 7³ = 74

or, 7^1-2x+6+3 = 7⁴

or, 1 – 2x = 4

or, 2x = 6

or, x = 3

(j) 27^x+1 = 9^x+1 × 81^x-2

or, (3)^3x+3 = (3)^2x+2 × (3)^4x-8

or, (3)^3x+3 =(3)^2x+2+4x-8

or, 3x + 3 = 6x – 6

or, 3x = 9

or, x = 3

(k) 25^x+1 = 5 × (125)^x-1

or, (5)^2x+2 = 5 (5)^3x-3

or, (5)^2x+2 = (5)^1+3x-3

or, 2x + 2 = 3x – 2

or, x = 4

(l) (36)^x × 6⁸ = (216)^3x-2

or, (6)^2x × (6)³ = (6)^9x-6

or, (6)^2x+8 = (6)

or, 2x + 8 = 9x – 6 or, 7x = 14

or, x = 2

(m) (125)^y-4 = (25)^y-2

or, (5)^3y – 12 = (5)^2y-4

or, 3y – 12 = 2y – 4

or, y = 8

(n) (27)⁷⁺¹ = (9)^y+3

or, (3)^3y+3 (3)^2y+6

or, 3y + 3 = 2y + 6

or, y = 3

(o) (2/5)^x-2 = (25/4)³

or, (2/5)^x-2 = (5/2)⁶

or, (2/5)^x-2 = (2/5)^-6

or, x – 2 = – 6

or, x = – 4

(p) (2/3)^x-8 = (8/27)^x

or, (3/2)^x-8 = (2/3)^3x

or, (3/2)^x-8 = (3/2)^-3x

or, x- 8 = – 3x

or, 4x = 8

or, x = 2

Question no – (2) 

Solution :

(a) (2/3)^-4 × (2/3)³

or, (2/3)^-12 = (2/3)^4m

or, 4m = – 12

or, m = – 3

(b) (2/7)^-17 ÷ (2/7)⁸ = (2/7)^2m+1

or, (2/7)^-17-8 = (2/7)^2m+1

or, – 25 = 2m + 1

or, 2m = – 26

or, m = – 13

(c) (3/5)^-3 × (3/5)¹⁰ = (3/5)^3m+1

or, (3/5)⁷ = (3/5)^3m+1

or, 3m + 1 = 7

or, 3m = 6

or, m = 2

(d) (4/5)^3m+1 × (4/5)^-15 = (4/5)^m

or, (4/5)^3m-14 = (4/5)^m

or, 3^{m – 14} = m

or, 2m = 14

or, m = 7

Question no – (3) 

Solution :

(a) 10⁶/5⁷ × 2⁵

= 5⁶ × 2⁶/5⁷ × 2⁵

= 2/5

(b) 6⁷/2⁵ × 3⁶

= 2⁷ × 3⁷/2⁵ ×3⁶

= 2² × 3/1

= 12

(c) 8² × 9³/6⁵

= 2⁶ × 3⁶/2⁵ × 3⁵

= 2 × 3

= 6

Question no – (4) 

Solution :

(a) [(2/3)^-2]³

= (2/3)^-6

= (3/2)⁶

= 729/64

(b) (2x^{– 3})⁴ × (x⁴)³

= 2x^-12 × x¹²

= 2

(c) 2^2n-1 × 4^1-n

= 2^2n-1 × 2^2n-1+2-2n

= 2¹

= 2

(d) (7³)⁴ × (7^-2)⁵

= 7¹² × 7^-10

= 7^12-10

= 7²

= 49

(e) (2³ × 3^-2)^-2

= 2^-6 × 3⁴

= 3⁴/2⁶

= 81/64

(f)  64ⁿ × (4)^2-3n

= (4)³ⁿ × (2)^2-3n

= (4)^3n+2-3n

= 4²

= 16

(g) (5^-2)⁴ × (125)²

= (5)^-8 × (53)² = 5^-8+6

= 5^-2

= 1/25

(h) (9^-5)² ÷ 27^-6

= 9^-10 ÷ 3^-18

= 3^-20 ÷ 3¹⁸

= 3^-2

= 1/9

Revision Exercise Questions Solution :  

Question no – (1)

Solution :

(a) (7/8)^-3 × (7/8)²

To evaluate the given problem,

= (7/8)^-3 × (7/8)²

= (7/8)^-3+2

= 8/7

(b) [(5/4)^-1 – (5/3)^-1] ÷ (7/5)^-2

To evaluate the given problem,

= [(5/4)^-1 – (5/3)^-1] ÷ (7/5)^-2

= [4/5 – 3/5] ÷ (5/7)²

= [1/5] × 49/25

= 49/125

Question no – (2) 

Solution :

(a) (2/3)^3x+1 × (2/3)³ = (2/3)¹⁰

We need to find,

= x = ?

= (2/3)^3x+1 × (2/3)³ = (2/3)¹⁰

= (2/3)^3x+1 × (2/3)³ = (2/3)¹⁰

or, (2/3)³ⁿ⁺⁴ = (2/3)³

or, 3x + 4 = 3

or, 3x = -1

or, x = -1/3

(b) (25)^x+2 = 5^x-4 × 125^x

Here we find the x

(25)^x+2 = 5^x-4 × 125^x

or, (5)^2x+4 = 5^x-4 × (5)^3x

or, (5)^2x+4 = (5)^4x-4

or, 2x + 4 = 4x – 4

or, 2x = 8

or, x = 4

Thus, here we find the x = 4

Question no – (3)

Solution :

(a) (-2/3)^-3 ÷ (4/27)^-2

To evaluate the given problem,

= (-2/3)^-3 ÷ (4/27)^-2

= (-3/2)³ ÷ (27/4)²

= – 27/8 × 16/721

= – 2/27

(b) (9/4)^-2 ÷ (2/3)³ + 4 (2/7)⁰

Now we evaluate the given problem by

= (9/4)^-2 ÷ (2/3)³ + 4 (2/7)⁰

= (4/9)² × (3/2)³ + 4.1

= 16/81 × 27/8 + 4

= 2/3 + 4

= 16/3

Question no – (4) 

Solution :

(a) (5)³ × (5)⁸ × (5)¹¹

= 5³⁺⁸⁺¹¹

= 5²²

(b) (- 6)⁴ × 3⁵ ÷ 2⁴

= (6)⁴ × 3⁵/2⁴ = 3⁴ × 2⁴ × 3⁵/2⁴

= 3⁴ × 3⁵ = 3⁹

(c) 3² × 2^-3 × (1/4)^-2

= 3² × 1/2³ × (4/1)²

= 3² × 1/2³ × 16

= 3² × 2

= 18

(d) (5)^-4 × (4/5)^-4 × 2⁸

= 1/5⁴ × 5⁴/4⁴ × 2⁸

= 2⁸/4⁴

= 2⁸/2⁸

= 1

Question no – (5)

Solution :

7²³ + 7²⁴ + 7²⁵ + 7²⁶/16

Now to simplify the given problem,

= 7²³ + 7²⁴ + 7²⁵ + 7²⁶/16

= 7²³ (1 + 7 + 7² + 7³)/16

= 7²³ (1 + 7 + 49 + 343)/16

= 7²³ (400)/16

= 7²³ × 25

Next Chapter Solution : 

👉 Chapter 3 👈