RD Sharma Solutions Class 7 Chapter 6 Exponents

Exercise 6.2 

Question 1:  Using laws of exponents, simplify and write the answer in exponential form

(i) 2³ × 2⁴ × 2⁵

(ii) 5¹² ÷ 5³

(iii) (7²)³

(iv) (3²)⁵ ÷ 3⁴

(v) 3⁷ × 2⁷

(vi) (5²¹ ÷ 5¹³) × 5⁷ 

Solution 1:

(i) 2³ × 2⁴ × 2⁵

According to the law of exponents: a^m × aⁿ × a^p = a^(m+n+p)

So, the equation is written as

= 2^{(3 + 4 + 5)}

= 2¹²

(ii) 5¹² ÷ 5³

According to the law of exponents: a^m ÷ aⁿ = a^m-n

So, the equation is written as

= 5^{12 – 3} 

= 5⁹

(iii) (7²)³

According to the law of exponents: (a^m)ⁿ = a^{m x n}

So the equation is written as

= 7^{2 x 3} 

= 7⁶     

(iv) (3²)⁵ ÷ 3⁴

According to the law of exponents: (a^m)ⁿ = a^{m x n}

So the equation is written as

= 3^{2 x 5} ÷ 3⁴ 

= 3 ¹⁰ ÷ 3⁴

According to the law of exponents: a^m ÷ aⁿ = a^m-n

= 3^{(10 – 4)} 

= 3⁶

(v) 3⁷ × 2⁷

According to the law of exponents: a^m × b^m = (a × b)^m

So the equation is written as

= (3 × 2)⁷ 

= 6⁷

(vi) (5²¹ ÷ 5¹³) × 5⁷

According to the law of exponents: a^m ÷ aⁿ = a^m-n

= 5^{(21 -13)} × 5⁷

So the equation is written as

= 5⁸ × 5⁷

According to the law of exponents: a^m × aⁿ = a^{(m +n)}

So the equation is written as

= 5⁽⁸⁺⁷⁾ 

= 5¹⁵

Question 2: Simplify and express each of the following in exponential form:

(i) {(2³)⁴ × 2⁸} ÷ 2¹²

(ii) (8² × 8⁴) ÷ 8³

(iii) (5⁷/5²)× 5³ 

(iv) (5⁴× x¹⁰y⁵)/ (5⁴ × x⁷y⁴) 

Solution 2:

(i) {(2³)⁴ × 2⁸} ÷ 2¹²

= {2¹² × 2⁸} ÷ 2¹² 

= 2^{(12 + 8)} ÷ 2¹²

= 2²⁰ ÷ 2¹² 

= 2 ^{(20 – 12)} 

=  2⁸

(ii) (8² × 8⁴) ÷ 8³

= 8^{(2 + 4)} ÷ 8³

= 8⁶ ÷ 8³

= 8^(6-3) 

= 8³ 

= (2³)³ 

= 2⁹

(iii) (5⁷/5²) × 5³

= 5^(7-2) × 5³

= 5⁵ × 5³

= 5^{(5 + 3)} 

= 5⁸

(iv) (5⁴× x¹⁰y⁵)/ (5⁴ × x⁷y⁴)

= (5^4-4× x^10-7y^5-4)

= 5⁰x³y¹ [since 5⁰ = 1]

= 1x³y

Question 3: Simplify and express each of the following in exponential form:

(i) {(3²)³ × 2⁶} × 5⁶

(ii)(x/y) ¹² × y²⁴ × (2³)⁴

(iii)(5/2)⁶ × (5/2)²

(iv) (2/3)⁵× (3/5)⁵ 

Solution 3:

(i) {(3²)³ × 2⁶} × 5⁶

= {3⁶ × 2⁶} × 5⁶

= 6⁶ × 5⁶ 

= 30⁶

(ii) (x/y)¹² × y²⁴ × (2³)⁴

= (x¹²/y¹²) × y²⁴ × 2¹²

= x¹² × y^24-12 × 2¹²

= x¹² × y¹² × 2¹²

= (2xy)¹²

(iii)(5/2)⁶ × (5/2)²

= (5/2)6+2

= (5/2)8

(iv) (2/3)⁵× (3/5)⁵ 

= (2/5)⁵

Question 4: Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3. 

Solution 4:

= 9 × 9 × 9 × 9 × 9

= (9)⁵ = (3²)⁵

= 3¹⁰

Question 5: Simplify and write each of the following in exponential form:

(i) (25)³ ÷ 5³

(ii) (81)⁵ ÷ (3²)⁵

(iii) 9⁸ × (x²)⁵/ (27)⁴ × (x³)²

(iv) 3² × 7⁸ × 13⁶/ 21² × 91³ 

Solution 5:

(i) (25)³ ÷ 5³

= (5²)³ ÷ 5³

= 5⁶ ÷ 5³ 

= 5^{6 – 3}

= 5³

(ii) (81)⁵ ÷ (3²)⁵

= (81)⁵ ÷ 3¹⁰                                     ->81 = 3⁴

= (3⁴)⁵ ÷ 3¹⁰ 

= 3²⁰ ÷ 3¹⁰

= 3^20-10 

= 3¹⁰

(iii) 9⁸ × (x²)⁵/ (27)⁴ × (x³)²

= (3²)⁸ × (x²)⁵/ (3³)⁴× (x³)²

= 3¹⁶ × x¹⁰/3¹² × x⁶

= 3^16-12 × x^10-6

= 3⁴ × x⁴

= (3x)⁴

(iv) (3² × 7⁸ × 13⁶)/ (21² × 91³)

= (3² × 7²7⁶ × 13⁶)/(21²× 13³ × 7³)

= (21² × 7⁶ × 13⁶)/(21²× 13³ × 7³)

= (7⁶ × 13⁶)/(13³ × 7³)

= 91⁶/91³

= 91^6-3

= 91³

Question 6: Simplify:

(i) (3⁵)¹¹ × (3¹⁵)⁴ – (3⁵)¹⁸ × (3⁵)⁵

(ii) (16 × 2ⁿ⁺¹ – 4 × 2ⁿ)/(16 × 2ⁿ⁺² – 2 × 2ⁿ⁺²)

(iii) (10 × 5ⁿ⁺¹ + 25 × 5ⁿ)/(3 × 5ⁿ⁺² + 10 × 5ⁿ⁺¹)

(iv) (16)⁷ ×(25)⁵× (81)³/(15)⁷ ×(24)⁵× (80)³ 

Solution 6:

(i) (3⁵)¹¹ × (3¹⁵)⁴ – (3⁵)¹⁸ × (3⁵)⁵

= (3)⁵⁵ × (3)⁶⁰ – (3)⁹⁰ × (3)²⁵

= 3 ⁵⁵⁺⁶⁰ – 3⁹⁰⁺²⁵

= 3¹¹⁵ – 3¹¹⁵

= 0

(ii) (16 × 2ⁿ⁺¹ – 4 × 2ⁿ)/(16 × 2ⁿ⁺² – 2 × 2ⁿ⁺²)

= (2⁴ × 2⁽ⁿ⁺¹⁾ -2² × 2ⁿ)/(2⁴ × 2⁽ⁿ⁺²⁾ -2ⁿ⁺¹ × 2²)

= 2² × 2^(n+3-2n)/)2²× 2^(n+4-2n+1)

= 2ⁿ × 2³ – 2ⁿ/ 2ⁿ × 2⁴ – 2ⁿ × 2

= 2ⁿ(2³ – 1)/ 2ⁿ(2⁴ – 1)

= 8 -1 /16 -2

= 7/14

= (1/2)

(iii) (10 × 5ⁿ⁺¹ + 25 × 5ⁿ)/(3 × 5ⁿ⁺² + 10 × 5ⁿ⁺¹)

= (10 × 5ⁿ⁺¹ + 5² × 5ⁿ)/(3 × 5ⁿ⁺² + (2 × 5) × 5ⁿ⁺¹)

= (10 × 5ⁿ⁺¹ + 5 × 5ⁿ⁺¹)/(3 × 5ⁿ⁺² + (2 × 5) × 5ⁿ⁺¹)

= 5ⁿ⁺¹ (10+5)/ 5ⁿ⁺¹ (10+15)

= 15/25

= (3/5)

(iv) (16)⁷ ×(25)⁵× (81)³/(15)⁷ ×(24)⁵× (80)³

= (16)⁷ × (5²)⁵× (3⁴)³/(3 × 5 )⁷ ×(3 × 8)⁵× (16 × 5)³

= (16)⁷ × (5²)⁵× (3⁴)³/3⁷ × 5⁷ × 3⁵ × 8⁵× 16³× 5³

= (16)⁷/ 8⁵ × 16 ³

= (16)⁴/8⁵

= (2 × 8)⁴/8⁵

= 2⁴/8

= (16/8)

= 2

Question 7: Find the values of n in each of the following:

(i) 5²ⁿ × 5³ = 5¹¹

(ii) 9 × 3ⁿ = 3⁷

(iii) 8 × 2ⁿ⁺² = 32 

(iv) 7²ⁿ⁺¹ ÷ 49 = 7³

(v) (3/2)⁴ ×  (3/2) ⁵ = (3/2)²ⁿ⁺¹

(vi) (2/3)¹⁰× {(3/2)²}⁵ = (2/3)^{2n – 2} 

Solution:

(i)  5²ⁿ × 5³ = 5¹¹

= 5²ⁿ⁺³ = 5¹¹

On equating the coefficients, we find:

2n + 3 = 11

⇒2n = 11- 3

⇒2n = 8

⇒ n = (8/2)

⇒ n = 4

(ii)   9 × 3ⁿ = 3⁷

= (3)² × 3ⁿ = 3⁷

= (3)²⁺ⁿ = 3⁷

On equating the coefficients, we find:

2 + n = 7

⇒ n = 7 – 2 = 5

(iii) 8 × 2ⁿ⁺² = 32

= (2)³ × 2ⁿ⁺² = (2)⁵      [since 2³ = 8 and 2⁵ = 32]

= (2)³⁺ⁿ⁺² = (2)⁵

On equating the coefficients, we find:

3 + n + 2 = 5

⇒ n + 5 = 5

⇒ n = 5 -5

⇒ n = 0

(iv) 7²ⁿ⁺¹ ÷ 49 = 7³

= 7²ⁿ⁺¹ ÷ 7² = 7³  [since 49 = 7²]

= 7^2n+1-2 = 7³

= 7^2n-1=7³

On equating the coefficients, we find:

2n – 1 = 3

⇒ 2n = 3 + 1

⇒ 2n = 4

⇒ n =4/2 =2

(v) (3/2)⁴ ×  (3/2) ⁵ = (3/2)²ⁿ⁺¹

= (3/2)⁴⁺⁵ = (3/2)²ⁿ⁺¹

= (3/2)⁹ = (3/2)²ⁿ⁺¹

On equating the coefficients, we find:

2n + 1 = 9

⇒ 2n = 9 – 1

⇒ 2n = 8

⇒ n =  8/2

=4

(vi) (2/3)¹⁰× {(3/2)²}⁵ = (2/3)^{2n – 2}

= (2/3)¹⁰ × (3/2)¹⁰ = (2/3)^{2n – 2}

= 2 ¹⁰ × 3¹⁰/3¹⁰ × 2¹⁰ = (2/3)^{2n – 2}

= 1 = (2/3)^{2n – 2}

= (2/3)⁰ = (2/3)^{2n – 2}

On equating the coefficients, we find:

0 =2n -2

2n -2 =0

2n =2

n = 1

Question 8: If (9ⁿ × 3² × 3ⁿ – (27)ⁿ)/ (3³)⁵ × 2³ = (1/27), find the value of n. 

Solution 8:

= (9ⁿ × 3² × 3ⁿ – (27)ⁿ)/ (3³)⁵ × 2³ = (1/27)

= (3²)ⁿ × 3³ × 3ⁿ – (3³)ⁿ/ (3¹⁵ × 2³) = (1/27)

= 3^(2n+2+n) – (3³)ⁿ/ (3¹⁵ × 2³) = (1/27)

= 3⁽³ⁿ⁺²⁾– (3³)ⁿ/ (3¹⁵ × 2³) = (1/27)

= 3³ⁿ × 3² – 3³ⁿ/ (3¹⁵ × 2³) = (1/27)

= 3³ⁿ × (3² – 1)/ (3¹⁵ × 2³) = (1/27)

= 3³ⁿ × (9 – 1)/ (3¹⁵ × 2³) = (1/27)

= 3³ⁿ × (8)/ (3¹⁵ × 2³) = (1/27)

= 3³ⁿ × 2³/ (3¹⁵ × 2³) = (1/27)

= 3³ⁿ/3¹⁵ = (1/27)

= 3^3n-15 = (1/27)

= 3^3n-15 = (1/3³)

= 3^3n-15 = 3^-3

On equating the coefficients, we find:

3n -15 = -3

⇒ 3n = -3 + 15

⇒ 3n = 12

⇒ n = 12/3 = 4 

Exercise 6.3

Question 1: Express the following numbers in the standard form:
(i) 3908.78
(ii) 5,00,00,000
(iii) 3,18,65,00,000
(iv) 846 × 10⁷
(v)723 × 10⁹ 

Solution 1:

(i) 3908.78

= 3.90878 × 10³ [Here the decimal point is moved left up to 3 places]

(ii) 5,00,00,000

= 5,00,00,000.00 = 5 × 10⁷ [Here, the decimal point is moved left up to 7 places]

(iii) 3,18,65,00,000

= 3,18,65,00,000.00

= 3.1865 × 10⁹ [here, the decimal point is moved left up to 9 places]

(iv) 846 × 10⁷

= 8.46 × 10² × 10 [here the decimal point is moved left up to 2 places]

= 8.46 × 10⁹ [since a^m × aⁿ = a^m+n]

(v) 723 × 10⁹

= 7.23 × 10² × 10⁹ [here the decimal point is moved left up to 2 places]

= 7.23 × 10¹¹ [ since a^m × aⁿ = a^m+n]

Question 2: Write the following numbers in the usual form:
(i) 4.83 × 10⁷
(ii) 3.21 × 10⁵
(iii) 3.5 × 10³

Solution 2:

(i) 4.83 × 10⁷

= 483 × 10^7-2 [hence the decimal point is moved right up to 2 places]

= 483 × 10⁵

= 4, 83, 00,000

(ii) 3.21 × 10⁵ 

= 321 × 10^5-2 [hence the decimal point is moved right up to 2 places]

= 321 × 10³

= 3, 21,000

(iii) 3.5 × 10³

= 35 × 10^3-1 [hence the decimal point is moved right up to 1 places]

= 35 × 10²

= 3,500

Question :3. Express the numbers appearing in the following statements in the standard form:

(i) The distance between the Earth and the Moon is 384,000,000 meters.
(ii) Diameter of the Earth is 1, 27, 56,000 meters.
(iii) Diameter of the Sun is 1,400,000,000 meters.
(iv) The universe is estimated to be about 12,000,000,000 years old. 

Solution:

(i) The distance between the Earth and the Moon is 384,000,000 meters.

Distance between the Earth and the Moon is 3.84 × 10⁸ meters.

[Here the decimal point is moved left up to 8 places.]

(ii) Diameter of the Earth is 1, 27, 56,000 meters.

Diameter of the Earth is 1.2756 × 10⁷ meters.

[Here the decimal point is moved left up to 7 places]

(iii) Diameter of the Sun is 1,400,000,000 meters.

Diameter of the Sun is 1.4 × 10⁹ meters.

[Here the decimal point is moved left up to 9 places]

(iv) The universe is estimated to be about 12,000,000,000 years old.

The universe is estimated to be about 1.2× 10¹⁰ years old.

[Here the decimal point is moved left up to 10 places] 

Exercise 6.4 

Question 1: Write the following numbers in the expanded exponential forms:
(i) 20068
(ii) 420719
(iii) 7805192
(iv) 5004132
(v) 927303 

Solution 1:

(i) 20068

= 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

(ii) 420719

= 4 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰

(iii) 7805192

 = 7 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 5 × 10³ + 1 × 10² + 9 × 10¹ + 2 × 10⁰

(iv) 5004132

= 5 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 4 × 10³ + 1 × 10² + 3 × 10¹ + 2 × 10⁰

(v) 927303

= 9 × 10⁵ + 2 × 10⁴ + 7 × 10³ + 3 × 10² + 0 × 10¹ + 3 × 10⁰

Question 2: Find the number from each of the following expanded forms:
(i) 7 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
(ii) 5 × 10⁵ + 4 × 10⁴ + 2 × 10³ + 3 × 10⁰
(iii) 9 × 10⁵ + 5 × 10² + 3 × 10¹
(iv) 3 × 10⁴ + 4 × 10² + 5 × 10⁰ 

Solution 2:

(i) 7 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰

= 7 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1

= 70000 + 6000 + 0 + 40 + 5

= 76045

(ii) 5 × 10⁵ + 4 × 10⁴ + 2 × 10³ + 3 × 10⁰

= 5 × 100000 + 4 × 10000 + 2 × 1000 + 3 × 1

= 500000 + 40000 + 2000 + 3

= 542003

(iii) 9 × 10⁵ + 5 × 10² + 3 × 10¹

= 9 × 100000 + 5 × 100 + 3 × 10

= 900000 + 500 + 30

= 900530

(iv) 3 × 10⁴ + 4 × 10² + 5 × 10⁰

= 3 × 10000 + 4 × 100 + 5 × 1

= 30000 + 400 + 5

= 30405