Class 9 RD Sharma Solutions Chapter 2 Exponents of Real Numbers Exercise 2.1 (original) (raw)

Last Updated : 6 Sep, 2024

Chapter 2 of Class 9 RD Sharma focuses on the concept of exponents of real numbers. This chapter introduces students to the fundamental principles of the exponents and their application in simplifying mathematical expressions. Understanding exponents is crucial for grasping more advanced topics in algebra and calculus.

Exponents of Real Numbers

Exponents also known as indices represent the power to which a number or variable is raised. They are essential in expressing large numbers in the compact form and in solving equations involving exponential terms. The general form is ab where a is the base and b is the exponent. The Exponents follow specific rules and properties such as the product rule, quotient rule, and power rule which are used to simplify and manipulate expressions involving powers.

Question 1 (i). Simplify 3(a4b3)10 × 5(a2b2)3

**Solution:

Given 3(a4b3)10 × 5(a2b2)3

= 3 × a40 × b30 × 5 × a6 × b6

= 3 × a46 × b36 × 5 [am × an = am+n]

= 15 × a46 × b36

= 15a46b36

Thus, 3(a4b3)10 × 5(a2b2)3 = 15a46b36

Question 1 (ii). Simplify (2x-2y3)3

**Solution:

Given (2x-2y3)3

= 23 × x-6 × y9

= 8 × x-6 × y9 [am × an = am+n]

= 8x-6y9

Thus, (2x-2y3)3 = 8x-6y9

Question 1 (iii). Simplify \frac{(4×10^7)(6×10^{-5})}{8×10^4}

**Solution:

Given \frac{(4×10^7)(6×10^{-5})}{8×10^4}

=\frac{4×10^7×6×10^{-5}}{8×10^4}

=\frac{24×10^{7+(-5)}}{8×10^4} [am × an = am+n]

=\frac{24×10^2}{8×10^4}

= 3/102

= 3/100

Thus, \frac{(4×10^7)(6×10^{-5})}{8×10^4} =3/100

Question 1 (iv). Simplify \frac{4ab^2(-5ab^3)}{10a^2b^2}

**Solution:

Given \frac{4ab^2(-5ab^3)}{10a^2b^2}

= \frac{4×a×b^2×(-5)×a×b^3}{10a^2b^2}

= \frac{-20×a×a×b^2×b^3}{10a^2b^2}

= \frac{-20×a^{1+1}×b^{2+3}}{10a^2b^2} [am × an = am+n]

= -2×a2×b5×a-2×b-2

= -2×a2+(-2)×b5+(-2) [am × an = am+n]

= -2×a0×b3

= -2b3 [a0=1]

Thus, \frac{4ab^2(-5ab^3)}{10a^2b^2} =-2b3

Question 1 (v). Simplify (\frac{x^2y^2}{a^2b^3})^n

**Solution:

Given(\frac{x^2y^2}{a^2b^3})^n

= \frac{(x^2)^n(y^2)^n}{(a^2)^n(b^3)^n}

= \frac{x^{2n}y^{2n}}{a^{2n}b^{3n}} [am × an = am+n]

Thus, (\frac{x^2y^2}{a^2b^3})^n =\frac{x^{2n}y^{2n}}{a^{2n}b^{3n}}

Question 1 (vi). Simplify \frac{(a^{3n-9})^6}{a^{2n-4}}

**Solution:

Given \frac{(a^{3n-9})^6}{a^{2n-4}}

= \frac{a^{6(3n-9)}}{a^{2n-4}} [(am)n = amn]

= \frac{a^{18n-54}}{a^{2n-4}}

= a18n-54 × a-(2n-4) [am × an = am+n]

= a18n-54-2n+4

= a16n-50

Thus, \frac{(a^{3n-9})^6}{a^{2n-4}} = a16n-50

Question 2 (i) If a = 3 and b = -2,find the value of aa + bb

**Solution:

Given a = 3 and b = -2

On substituting the value of a and b in aa + bb, we get

aa + bb = 33 + (-2)-2

= 27 + 1/4

= (108 + 1)/4

= 109/4

Thus, aa + bb = 109/4

Question 2 (ii). If a = 3 and b = -2,find the value of ab + ba

**Solution:

Given a = 3 and b = -2

On substituting the value of a and b in ab + ba, we get

ab + ba = 3-2 + (-2)3

= 1/9 + (-8)

= (1 - 72)/9

= -71/9

Thus, ab + ba = -71/9

Question 2 (iii). If a = 3 and b = -2,find the value of (a + b)ab.

**Solution:

Given a = 3 and b = -2

On substituting the value of a and b in (a + b)ab, we get

(a + b)ab = (3 + (-2))3×-2

= (1)-6

= 1

Thus, (a + b)ab = 1

Question 3 (i). Prove that (\frac{x^a}{x^b})^{a^2+ab+b^2}×(\frac{x^b}{x^c})^{b^2+bc+c^2}×(\frac{x^c}{x^a})^{c^2+ca+a^2}=1

**Solution:

Let us first solve left-hand side of the given equation

(\frac{x^a}{x^b})^{a^2+ab+b^2}×(\frac{x^b}{x^c})^{b^2+bc+c^2}×(\frac{x^c}{x^a})^{c^2+ca+a^2}

By using the formula (am)n = amn, we get

= \frac{x^{a(a^2+ab+b^2)}}{x^{b(a^2+ab+b^2)}}×\frac{x^{b(b^2+bc+c^2)}}{x^{c(b^2+bc+c^2)}}×\frac{x^{c(c^2+ca+a^2)}}{x^{a(c^2+ca+a^2)}}

By using the formula am/an = am-n, we get

= x^{a(a^2+ab+b^2)-b(a^2+ab+b^2)}×x^{b(b^2+bc+c^2)-c(b^2+bc+c^2)}×x^{c(c^2+ca+a^2)-a(c^2+ca+a^2)}

= x^{(a-b)(a^2+ab+b^2)}×x^{(b-c)(b^2+bc+c^2)}×x^{(c-a)(c^2+ca+a^2)}

= x^{(a^3-b^3)}×x^{(b^3-c^3)}×x^{(c^3-a^3)}

By using the formula am × an = am+n , we get

=x^{(a^3-b^3+b^3-c^3+c^3-a^3)}

= x

= 1

= Right-hand side of the given equation

Thus, we proved that (\frac{x^a}{x^b})^{a^2+ab+b^2}×(\frac{x^b}{x^c})^{b^2+bc+c^2}×(\frac{x^c}{x^a})^{c^2+ca+a^2}=1

Question 3 (ii). Prove that (\frac{x^a}{x^b})^c×(\frac{x^b}{x^c})^a×(\frac{x^c}{x^a})^b=1

**Solution:

Let us consider the left-hand side of the given equation

(\frac{x^a}{x^b})^c×(\frac{x^b}{x^c})^a×(\frac{x^c}{x^a})^b

By using the formula, (am)n = amn, we get

= \frac{x^{ac}}{x^{bc}}×\frac{x^{ba}}{x^{ca}}×\frac{x^{cb}}{x^{ab}}

= \frac{x^{ac}×x^{ba}×x^{cb}}{x^{bc}×x^{ca}×x^{ab}}

= \frac{x^{ac+ba+bc}}{x^{bc+ca+ab}} [am × an = am+n]

= 1

= Right-hand side of the given equation

Thus, we proved that (\frac{x^a}{x^b})^c×(\frac{x^b}{x^c})^a×(\frac{x^c}{x^a})^b=1

Question 3 (iii). Prove that (\frac{x^a}{x^{-b}})^{a^2-ab+b^2}×(\frac{x^b}{x^{-c}})^{b^2-bc+c^2}×(\frac{x^c}{x^{-a}})^{c^2-ca+a^2}=x^{2(a^3+b^3+c^3)}

**Solution:

Let us first solve left-hand side of the given equation

(\frac{x^a}{x^{-b}})^{a^2-ab+b^2}×(\frac{x^b}{x^{-c}})^{b^2-bc+c^2}×(\frac{x^c}{x^{-a}})^{c^2-ca+a^2}=1

By using the formula (am)n = amn, we get

= \frac{x^{a(a^2-ab+b^2)}}{x^{-b(a^2-ab+b^2)}}×\frac{x^{b(b^2-bc+c^2)}}{x^{-c(b^2-bc+c^2)}}×\frac{x^{c(c^2-ca+a^2)}}{x^{-a(c^2-ca+a^2)}}

By using the formula am/an = am-n, we get

= x^{a(a^2-ab+b^2)+b(a^2-ab+b^2)}×x^{b(b^2-bc+c^2)+c(b^2-bc+c^2)}×x^{c(c^2-ca+a^2)+a(c^2-ca+a^2)}

= x^{(a+b)(a^2-ab+b^2)}×x^{(b+c)(b^2-bc+c^2)}×x^{(c+a)(c^2-ca+a^2)}

= x^{(a^3+b^3)}×x^{(b^3+c^3)}×x^{(c^3+a^3)}

By using the formula am × an = am+n , we get

= x^{(a^3+b^3+b^3+c^3+c^3+a^3)}

= x^{2(a^3+b^3+c^3)}

= Right-hand side of the given equation

Thus, we proved that (\frac{x^a}{x^{-b}})^{a^2-ab+b^2}×(\frac{x^b}{x^{-c}})^{b^2-bc+c^2}×(\frac{x^c}{x^{-a}})^{c^2-ca+a^2}=x^{2(a^3+b^3+c^3)}

Question 4 (i). Prove that \frac1{1+x^{a-b}}+\frac1{1+x^{b-a}}=1

**Solution:

Let us first consider the left-hand side of given equation

\frac1{1+x^{a-b}}+\frac1{1+x^{b-a}}

= \frac1{1+\frac{x^a}{x^b}}+\frac1{1+\frac{x^b}{x^a}}

= \frac1{\frac{x^b+x^a}{x^b}}+\frac1{\frac{x^a+x^b}{x^a}}

= \frac{x^b}{x^a+x^b}+\frac{x^a}{x^a+x^b}

= \frac{x^a+x^b}{x^b+x^a}

= 1

= Right-hand side of the given equation

Thus, we proved that \frac1{1+x^{a-b}}+\frac1{1+x^{b-a}}=1

Question 4 (ii). Prove that \frac1{1+x^{b-a}+x^{c-a}}+\frac1{1+x^{a-b}+x^{c-b}}+\frac1{1+x^{b-c}+x^{c-a}}=1

**Solution:

Let us first consider the left-hand side of given equation

\frac1{1+x^{b-a}+x^{c-a}}+\frac1{1+x^{a-b}+x^{c-b}}+\frac1{1+x^{b-c}+x^{c-a}}

= \frac1{1+\frac{x^b}{x^a}+\frac{x^c}{x^a}}+\frac1{1+\frac{x^a}{x^b}+\frac{x^c}{x^b}}+\frac1{1+\frac{x^b}{x^c}+\frac{x^a}{x^c}}

= \frac1{\frac{x^a+x^b+x^c}{x^a}}+\frac1{\frac{x^a+x^b+x^c}{x^b}}+\frac1{\frac{x^a+x^b+x^c}{x^c}}

= \frac{x^a}{x^a+x^b+x^c}+\frac{x^b}{x^a+x^b+x^c}+\frac{x^c}{x^a+x^b+x^c}

= \frac{x^a+x^b+x^c}{x^a+x^b+x^c}

= 1

= Right-hand side of the given equation

Thus, we proved that \frac1{1+x^{b-a}+x^{c-a}}+\frac1{1+x^{a-b}+x^{c-b}}+\frac1{1+x^{b-c}+x^{c-a}}=1

Question 5 (i). Prove that \frac{a+b+c}{a^{-1}b^{-1}+b^{-1}c^{-1}+a^{-1}c^{-1}}=abc

**Solution:

Let us first consider the left-hand side of given equation

\frac{a+b+c}{a^{-1}b^{-1}+b^{-1}c^{-1}+a^{-1}c^{-1}}

= \frac{a+b+c}{\frac1{ab}+\frac1{bc}+\frac1{ca}}

= \frac{a+b+c}{\frac{a+b+c}{abc}}

= abc

= Right hand side of the given equation

Thus, we proved that \frac{a+b+c}{a^{-1}b^{-1}+b^{-1}c^{-1}+a^{-1}c^{-1}}=abc

Question 5 (ii). Prove that (a^{-1}+b^{-1})^{-1}=\frac{ab}{a+b}

**Solution:

Let us first consider the left hand side of given equation

(a^{-1}+b^{-1})^{-1}

= \frac1{a^{-1}+b^{-1}}

= \frac1{\frac1a+\frac1b}

= \frac1{\frac{b+a}{ab}}

= \frac{ab}{a+b}

= Right hand side of the given equation

Thus, we proved that (a^{-1}+b^{-1})^{-1}=\frac{ab}{a+b}

Question 6. If abc = 1, show that \frac1{1+a+b^{-1}}+\frac1{1+b+c^{-1}}+\frac1{1+c+a^{-1}}=1

**Solution:

Given abc = 1

⇒ c = 1/ab

Let us first consider the left-hand side of given equation

\frac1{1+a+b^{-1}}+\frac1{1+b+c^{-1}}+\frac1{1+c+a^{-1}}

= \frac1{1+a+\frac1b}+\frac1{1+b+\frac1c}+\frac1{1+c+\frac1a}

= \frac1{\frac{b+ab+1}{b}}+\frac1{\frac{c+bc+1}c}+\frac1{\frac{a+ac+1}a}

= \frac{b}{b+ab+1}+\frac{c}{c+bc+1}+\frac{a}{a+ac+1}

By substituting the value of c in above equation, we get

= \frac{b}{b+ab+1}+\frac{\frac1{ab}}{\frac1{ab}+b(\frac1{ab})+1}+\frac{a}{a+a(\frac1{ab})+1}

= \frac{b}{b+ab+1}+\frac{\frac1{ab}}{\frac1{ab}+\frac{b}{ab}+\frac{ab}{ab}}+\frac{a}{\frac{ab}{b}+\frac1{b}+\frac{b}{b}}

= \frac{b}{b+ab+1}+\frac{\frac1{ab}}{\frac{1+b+ab}{ab}}+\frac{a}{\frac{ab+1+b}{b}}

= \frac{b}{b+ab+1}+\frac{\frac{ab}{ab}}{1+b+ab}+\frac{ab}{ab+b+1}

= \frac{b+1+ab}{b+ab+1}

= 1

= Right hand side of the given equation

Thus, we have shown that if abc = 1, \frac1{1+a+b^{-1}}+\frac1{1+b+c^{-1}}+\frac1{1+c+a^{-1}}=1

Question 7 (i). Simplify \frac{3^n×9^{n+1}}{3^{n-1}×9^{n-1}}

**Solution:

Given \frac{3^n×9^{n+1}}{3^{n-1}×9^{n-1}}

= \frac{3^n×(3^2)^{n+1}}{3^{n-1}×(3^2)^{n-1}}

= \frac{3^n×3^{2n+2}}{3^{n-1}×3^{2n-2}}

= \frac{3^{n+2n+2}}{3^{n-1+2n-2}} [am × an = am+n]

= \frac{3^{3n+2}}{3^{3n-3}}

= 33n+2-(3n-3) [am/an = am-n]

= 35

= 243

Thus, \frac{3^n×9^{n+1}}{3^{n-1}×9^{n-1}} = 243

Question 7 (ii). Simplify \frac{5×25^{n+1}-25×5^{2n}}{5×5^{2n+3}-25^{n+1}}

**Solution:

Given \frac{5×25^{n+1}-25×5^{2n}}{5×5^{2n+3}-25^{n+1}}

= \frac{5×(5^2)^{n+1}-(5^2)×5^{2n}}{5×5^{2n+3}-(5^2)^{n+1}}

= \frac{5×(5^{2n+2})-(5^2)×5^{2n}}{5×5^{2n+3}-5^{2n+2}}

= \frac{5^{1+2n+2}-5^{2+2n}}{5^{1+2n+3}-5^{2n+2}} [am × an = am+n]

= \frac{5^{2+2n}(5-1)}{5^{2+2n}(5^2-1)}

= 4/24

= 1/6

Thus, \frac{5×25^{n+1}-25×5^{2n}}{5×5^{2n+3}-25^{n+1}} = 1/6

Question 7 (iii). Simplify \frac{5^{n+3}-6×5^{n+1}}{9×5^n-2^2×5^n}

**Solution:

Given, \frac{5^{n+3}-6×5^{n+1}}{9×5^n-2^2×5^n}

= \frac{5^{n+1}(5^2-6)}{5^n(9-2^2)}

=\frac{5^n×5×(25-6)}{5^n(9-4)}

= (19 × 5)/5

= 19

Thus, \frac{5^{n+3}-6×5^{n+1}}{9×5^n-2^2×5^n}=19

Question 7 (iv). Simplify \frac{6(8)^{n+1}+16(2)^{3n-2}}{10(2)^{3n+1}-7(8)^n}

**Solution:

Given \frac{6(8)^{n+1}+16(2)^{3n-2}}{10(2)^{3n+1}-7(8)^n}

= \frac{6(2^3)^{n+1}+16(2)^{3n-2}}{10(2)^{3n+1}-7(2^3)^n}

= \frac{6(2^{3n+3})+16(2)^{3n-2}}{10(2)^{3n+1}-7(2^{3n})}

= \frac{6×2^{3n}(2^3)+16(2^{3n})2^{-2}}{10(2)^{3n}(2^1)-7(2^{3n})}

= \frac{2^{3n}((6×2^3)+(16×\frac1{2^2}))}{2^{3n}((10×2)-7)}

= \frac{6×8+(16×\frac14)}{20-7}

= (48 + 4)/13

= 52/13

= 4

Thus, \frac{6(8)^{n+1}+16(2)^{3n-2}}{10(2)^{3n+1}-7(8)^n}=4

Question 8 (i). Solve the equation 72x+3 = 1 for x.

**Solution:

Given equation 72x+3 = 1

We know that, for any a∈ Real numbers, a0 = 1

Let a = 7

⇒ 72x+3 = 70

Since the bases are equal, let us equate the exponents

⇒ 2x + 3 = 0

⇒ x = -3/2

Thus, the value of x is -3/2

Question 8 (ii). Solve the equation 2x+1 = 4x-3 for x.

**Solution:

Given 2x+1 = 4x-3

We can write 4 = 22

⇒ 2x+1 = 22(x-3)

⇒ 2x+1 = 22x-6

Since the bases are equal, let us equate the exponents

⇒ x + 1 = 2x - 6

⇒ x = 7

Thus, the value of x is 7

Question 8 (iii). Solve the equation 25x+3 = 8x+3 for x.

**Solution:

Given 25x+3 = 8x+3

We know that 8 = 23

⇒ 25x+3 = 23(x+3)

⇒ 25x+3 = 23x+9

Since the bases are equal, let us equate the exponents

⇒ 5x + 3 = 3x + 9

⇒ 5x - 3x = 9 - 3

⇒ 2x = 6

⇒ x = 3

Thus, the value of x is 3

Question 8 (iv). Solve the equation 42x = 1/32 for x.

**Solution:

Given 42x = 1/32

⇒ 22(2x) = 1/32

⇒ 22(2x) × 32 = 1

⇒ 24x × 25 = 1

⇒ 24x+5 = 20

Since the bases are equal, let us equate the exponents

⇒ 4x + 5 = 0

⇒ x = -5/4

Thus, the value of x is -5/4

Question 8 (v). Solve the equation 4x - 1 × (0.5)3-2x = (1/8)x for x.

**Solution:

Given 4x - 1 × (0.5)3-2x = (1/8)x

⇒ (2^2)^{x-1}×(\frac12)^{3-2x}=(\frac1{2^3})^x

⇒ (2^2)^{x-1}×(2^{-1})^{3-2x}=(2^{-3})^x

⇒ 22(x-1) × 2-(3-2x) = 2-3x

⇒ 22x-2-3+2x = 2-3x

⇒ 24x-5 = 2-3x

Since the bases are equal, let us equate the exponents

⇒ 4x - 5 = -3x

⇒ 7x = 5

⇒ x = 5/7

Thus, the value of x is 5/7

Question 8 (vi). Solve the equation 23x-7 = 256 for x.

**Solution:

Given 23x-7 = 256

⇒ 23x-7 = 28

Since the bases are equal, let us equate the exponents

⇒ 3x - 7 = 8

⇒ x = 15/3

⇒ x = 5

Thus, the value of x is 5

Question 9 (i). Solve the equation 22x - 2x+3 + 24 = 0 for x.

**Solution:

Given 22x - 2x+3 + 24 = 0

⇒ (2x)2 - 2 × 2x × 22 + (22)2 = 0

⇒ (2x - 22)2 = 0

⇒ 2x - 22 = 0

⇒ 2x = 22

Since the bases are equal, let us equate the exponents

⇒ x = 2

Thus, the value of x is 2

Question 9 (ii). Solve the equation 32x+4 + 1 = 2.3x+2 for x.

**Solution:

Given 32x+4 + 1 = 2.3x+2

⇒ (3^{x+2})^2+1=2.3^{x+2}

⇒ (3^{x+2})^2-2.3^{x+2}+1=0

⇒ (3x+2 - 1)2 = 0

⇒ 3x+2 - 1 = 0

⇒ 3x+2 = 30

Since the bases are equal, let us equate the exponents

⇒ x + 2 = 0

⇒ x = -2

Thus, the value of x is -2

Question 10. If 49392 = a4b2c3, find the values of a, b, and c where a, b and c are different positive primes.

**Solution:

Let us first find out prime factorization of 49392

Thus, 49392 = 24 × 32 × 73

Where 2, 3 and 7 are positive primes

49392 = 243273 = a4b2c3

Thus, on comparing, we get

a = 2,b = 3 and c = 7

Thus, the values of a, b and c are 2, 3, 7 respectively.

Question 11. If 1176 = 2a3b7c, find a, b and c.

**Solution:

Given 1176 = 2a3b7c

Let us first find out prime factorization of 1176

Thus, 1176 = 23 × 31 × 72

1176 = 233172 = 2a3b7c

Thus, on comparing, we get

a = 3, b = 1, c = 2

Thus, the values of a, b and c are 3, 1, 2 respectively.

Question 12. Given 4725 = 3a5b7c, find

(i) the integral values of a, b and c

(ii) the value of 2-a3b7c

**Solution:

Given 4725 = 3a5b7c

****(i)** Let us first find out prime factorization of 4725

Thus, 4725 = 33 × 52 × 71

4725 = 335271 = 3a5b7c

Thus, on comparing, we get

a = 3,b = 2,c = 1

Thus, the values of a, b and c are 3,2,1 respectively.

****(ii)** Here a = 3, b = 2, c = 1

On substituting these values in 2-a3b7c

2-a3b7c= 2-3×32×71

= 1/8 × 9 × 7 = 63/8

Thus, the value of 2-a3b7c is 63/8

Question 13. If a = xyp-1, b = xyq-1, c = xyr-1, prove that aq-rbr-pcp-q = 1.

**Solution:

Given a = xyp-1, b = xyq-1, c = xyr-1

aq-rbr-pcp-q=(xy^{p-1})^{q-r}(xy^{q-1})^{r-p}(xy^{r-1})^{p-q}

= x^{(q-r)}y^{(p-1)(q-r)}x^{(r-p)}y^{(r-p)(q-1)}x^{(p-q)}y^{(p-q)(r-1)}

= xq-r+r-p+p-q y(p-1)(q-r)+(r-p)(q-1)+(p-q)(r-1)

= xq-r+r-p+p-q ypq-q-pr+r+rq-r-pq+p+pr-p-qr+q

= x0y0

= 1

Thus, we proved that aq-rbr-pcp-q = 1

Conclusion

Mastering exponents is foundational for the advancing in algebra and higher mathematics. By understanding the rules and properties of the exponents students can efficiently simplify expressions and solve complex equations. This knowledge not only aids in the mathematical problem-solving but also in applying these concepts in the real-world scenarios and other scientific disciplines.