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Class 9 RD Sharma Solutions – Chapter 2 Exponents of Real Numbers- Exercise 2.1

  • Last Updated : 04 May, 2021

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

Solution:

Given 3(a4b3)10 × 5(a2b2)3

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= 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




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