Class 9 RD Sharma Solutions – Chapter 2 Exponents of Real Numbers- Exercise 2.2 | Set 2

Last Updated : 30 Apr, 2021

Question 12. Determine (8x)^x, if 9^x+2 = 240 + 9^x.

Solution:

We have,

=> 9^x+2 = 240 + 9^x

=> 9^x+2 − 9^x = 240

=> 9^x (9²− 1) = 240

=> 9^x = 240/80

=> 3^2x = 3

=> 2x = 1

=> x = 1/2

Therefore, (8x)^x = [8 × (1/2)]^1/2

= 4^1/2

= 2

Question 13. If 3^x+1 = 9^x−2, find the value of 2^1+x.

Solution:

We have,

=> 3^x+1 = 9^x−2

=> 3^x+1 = (3²)^x−2

=> 3^x+1 = 3^2x−4

=> x + 1 = 2x − 4

=> x = 5

Therefore, 2^1+x = 2¹⁺⁵

= 2⁶

= 64

Question 14. If 3^4x= (81)⁻¹ and (10)^1/y = 0.0001, find the value of 2^−x+4y.

Solution:

We are given,

=> 3^4x = (81)⁻¹

=> 3^4x = (3⁴)⁻¹

=> 3^4x = (3)⁻⁴

=> 4x = −4

=> x = −1

And also, (10)^1/y= 0.0001

=> (10)^1/y = (10)⁻⁴

=> 1/y = −4

=> y = −1/4

Therefore, 2^−x+4y= 2^1+4(−1/4)

= 2¹⁻¹

= 1

Question 15. If 5^3x = 125 and 10^y = 0.001. Find x and y.

Solution:

We are given,

=> 5^3x = 125

=> 5^3x = 5³

=> 3x = 3

=> x =1

Also, (10)^y = 0.001

=> 10^y = 10⁻³

=> y = −3

Therefore, the value of x is 1 and the value of y is –3.

Question 16. Solve the following equations:

(i) 3^x+1 = 27 × 3⁴

Solution:

We have,

=> 3^x+1 = 27 × 3⁴

=> 3^x+1 = 3³ × 3⁴

=> 3^x+1= 3⁷

=> x + 1 = 7

=> x = 6

(ii) $4^{2x}=\left(\sqrt[3]{16}\right)^{\frac{-6}{y}}=(\sqrt{8})^2$

Solution:

We have,

=> $4^{2x}=\left(\sqrt[3]{16}\right)^{\frac{-6}{y}}=(\sqrt{8})^2$

=> $(2^2)^{2x}=\left(\sqrt[3]{2^4}\right)^{\frac{-6}{y}}=(\sqrt{2^3})^2$

=> $(2)^{4x}=\left(2^\frac{4}{3}\right)^{\frac{-6}{y}}=(2^\frac{3}{2})^2$

=> $(2)^{4x}=(2)^{\frac{-8}{y}}=2^3$

=> 4x = −8/y = 3

=> x = 3/4 and y = −8/3

(iii) 3^x−1 × 5^2y−3 = 225

Solution:

We have,

=> 3^x−1 × 5^2y−3 = 225

=> 3^x−1 × 5^2y−3 = 3² × 5²

=> x − 1 = 2 and 2y − 3 = 2

=> x = 3 and 2y = 5

=> x = 3 and y = 5/2

(iv) 8^x+1 = 16^y+2 and (1/2)^3+x = (1/4)^3y

Solution:

We have,

=> 8^x+1 = 16^y+2

=> (2³)^x+1 = (2⁴)^y+2

=> 2^3x+3 = 2^4y+8

=> 3x + 3 = 4y + 8 . . . . (1)

Also, (1/2)^3+x = (1/4)^3y

=> (1/2)^3+x = [(1/2)²]^3y

=> (1/2)^3+x = (1/2)^6y

=> 3 + x = 6y

=> x = 6y − 3 . . . . (2)

Putting (2) in (1), we get,

=> 3(6y − 3) + 3 = 4y + 8

=> 18y − 9 + 3 = 4y + 8

=> 14y = 14

=> y = 1

Putting y = 1 in (2), we get,

x = 6(1) − 3 = 6 − 3 = 3

Therefore, the value of x is 1 and the value of y is –3.

(v) 4^x−1 × (0.5)^3−2x = (1/8)^x

Solution:

We have,

=> 4^x−1 × (0.5)^3−2x = (1/8)^x

=> (2²)^x−1× (1/2)^3−2x = [(1/2)³]^x

=> 2^2x−2 × 2^2x−3 = 2^−3x

=> 2^{2x−2+2x−3} = 2^−3x

=> 2^4x−5 = 2^−3x

=> 4x − 5 = −3x

=> 7x = 5

=> x = 5/7

(vi) $\sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2x}$

Solution:

We have,

=> $\sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2x}$

=> $\left(\frac{a}{b}\right)^{\frac{1}{2}}=\left(\frac{a}{b}\right)^{2x-1}$

=> 1/2 = 2x − 1

=> 2x = 3/2

=> x = 3/4

Question: 17. If a and b are distinct positive primes such that, $\sqrt[3]{a^6b^{-4}}=a^xb^{2y}$ find x and y.

Solution:

We have,

=> $\sqrt[3]{a^6b^{-4}}=a^xb^{2y}$

=> (a⁶b⁻⁴)^1/3 = a^xb^2y

=> a^6/3b^−4/3 = a^xb^2y

=> a²b^−4/3 = a^xb^2y

=> x = 2 and 2y = −4/3

=> x = 2 and y = −2/3

Question 18. If a and b are different positive primes such that,

(i) $\left(\frac{a^{-1}b^{2}}{a^2b^{-4}}\right)^7÷\frac{a^{3}b^{-5}}{a^{-2}b^{3}}=a^xb^y$ , find x and y.

Solution:

We have,

=> $\left(\frac{a^{-1}b^{2}}{a^2b^{-4}}\right)^7÷\frac{a^{3}b^{-5}}{a^{-2}b^{3}}=a^xb^y$

=> (a⁻¹⁻²b²⁺⁴)⁷ ÷ (a³⁺²b⁻⁵⁻³) = a^xb^y

=> (a⁻³b⁶)⁷ ÷ (a⁵b⁻⁸) = a^xb^y

=> (a⁻²¹b⁴²) ÷ (a⁵b⁻⁸) = a^xb^y

=> (a⁻²¹⁻⁵b⁴²⁺⁸) = a^xb^y

=> (a⁻²⁶b⁵⁰) = a^xb^y

=> x = −26, y = 50

(ii) (a + b)⁻¹(a⁻¹ + b⁻¹) = a^xb^y, find x+y+2.

Solution:

We have,

=> (a + b)⁻¹(a⁻¹ + b⁻¹) = a^xb^y

=> $(\frac{1}{a+b})(\frac{1}{a}+\frac{1}{b})$ = a^xb^y

=> $(\frac{1}{a+b})(\frac{b+a}{ab})$ = a^xb^y

=> 1/ab = a^xb^y

=> a⁻¹b⁻¹= a^xb^y

=> x = −1 and y = −1

So, x+y+2 = −1−1+2 = 0.

Question 19. If 2^x × 3^y × 5^z = 2160, find x, y and z. Hence compute the value of 3^x × 2^−y × 5^−z.

Solution:

We are given,

=> 2^x × 3^y× 5^z = 2160

=> 2^x× 3^y × 5^z = 2⁴× 3³× 5¹

=> x = 4, y = 3, z = 1

Therefore, 3^x × 2^−y × 5^−z = 3⁴ × 2⁻³ × 5⁻¹

= (81) (1/8) (1/5)

= 81/40

Question 20. If 1176 = 2^a × 3^b × 7^c, find the values of a, b and c. Hence, compute the value of 2^a × 3^b × 7^-c as a fraction.

Solution:

We are given,

=> 1176 = 2^a × 3^b × 7^c

=> 2³ × 3¹ × 7²= 2a × 3b × 7c

=> a = 3, b = 1, c = 2

Therefore, 2^a × 3^b × 7^−c = 2³ × 3¹ × 7⁻²

= (8) (3) (1/49)

= 24/49

Question 21. Simplify

(i) $\left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c}\left(\frac{x^{c+a}}{x^b}\right)^{c-a}$

Solution:

We have,

= $\left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c}\left(\frac{x^{c+a}}{x^b}\right)^{c-a}$

= (x^a+b−c)^a−b(x^b+c−a)^b−c(x^c+a−b)^c−a

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

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

= x⁰

= 1

(ii) $\sqrt[lm]{\frac{x^l}{x^m}}×\sqrt[mn]{\frac{x^m}{x^n}}×\sqrt[nl]{\frac{x^n}{x^l}}$

Solution:

We have,

=> $\sqrt[lm]{\frac{x^l}{x^m}}×\sqrt[mn]{\frac{x^m}{x^n}}×\sqrt[nl]{\frac{x^n}{x^l}}$

=> $(x^{l-m})^{\frac{1}{lm}}×(x^{m-n})^{\frac{1}{mn}}×(x^{n-l})^{\frac{1}{nl}}$

=> $x^{\frac{l-m}{lm}}×x^{\frac{m-n}{mn}}×x^{\frac{n-l}{nl}}$

=> $x^{\frac{l-m}{lm}+\frac{m-n}{mn}+\frac{n-l}{nl}}$

=> $x^{\frac{nl-mn+ml-nl+mn-ml}{mnl}}$

=> $x^{\frac{0}{mnl}}$

=> x⁰

= 1

Question 22. Show that $\frac{(a+\frac{1}{b})^m×(a-\frac{1}{b})^n}{(b+\frac{1}{a})^m×(b-\frac{1}{a})^n}=(\frac{a}{b})^{m+n}$ .

Solution:

We have,

L.H.S. = $\frac{(a+\frac{1}{b})^m×(a-\frac{1}{b})^n}{(b+\frac{1}{a})^m×(b-\frac{1}{a})^n}$

= $\frac{(\frac{ab+1}{b})^m×(\frac{ab-1}{b})^n}{(\frac{ab+1}{a})^m×(\frac{ab-1}{a})^n}$

= $(\frac{a}{b})^m×(\frac{a}{b})^n$

= $(\frac{a}{b})^{m+n}$

= R.H.S.

Hence proved.

Question 23. (i) If a = x^m+ny^l, b = x^n+ly^m and c = x^l+myⁿ, prove that a^m−nb^n−lc^l−m = 1.

Solution:

Given, a = x^m+ny^l, b = x^n+ly^m and c = x^l+myⁿ.

We have,

L.H.S. = a^m−n b^n−l c^l−m

= (x^m+ny^l)^m−n(x^n+ly^m)^n−l(x^l+myⁿ)^l−m

= $(x^{m^2-n^2}y^{lm-ln})(x^{n^2-l^2}y^{mn-ml})(x^{l^2-m^2}y^{nl-mn})$

= $x^{m^2-n^2+n^2-l^2+l^2-m^2}y^{lm-ln+mn-ml+nl-mn}$

= x⁰y⁰

= 1

= R.H.S.

Hence proved.

(ii) If x = a^m+n, y = a^n+land z = a^l+m, prove that x^myⁿz^l = xⁿy^lz^m.

Solution:

Given, x = a^m+n, y = a^n+l and z = a^l+m.

We have,

L.H.S. = x^myⁿz^l

= (a^m+n)^m(a^n+l)ⁿ (a^l+m)^l

= $a^{m^2+mn+n^2+ln+l^2+ml}$

= $a^{mn+n^2}×a^{nl+l^2}×a^{lm+m^2}$

= (a^m+n)ⁿ (a^n+l)^l (a^l+m)^m

= xⁿy^lz^m

= R.H.S.

Hence proved.

Suggest improvement

Class 9 RD Sharma Solutions - Chapter 2 Exponents of Real Numbers- Exercise 2.2 | Set 1

Class 9 RD Sharma Solutions - Chapter 3 Rationalisation- Exercise 3.1

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Chapter 1: Number Systems

Chapter 2: Exponents and Powers of Real Numbers

Chapter 3: Rationalisation

Chapter 4: Algebraic Identities

Chapter 5: Factorization of Algebraic Expressions

Chapter 6: Factorization of Polynomials

Chapter 7: Introduction to Euclidâ€™s Geometry

Chapter 8: Lines and Angles

Chapter 9: Triangle and Its Angles

Chapter 10: Congruent Triangles

Chapter 11: Coordinate Geometry

Chapter 12: Heronâ€™s Formula

Chapter 13: Linear Equations in Two Variables

Chapter 14: Quadrilaterals

Chapter 15: Areas of Parallelograms and Triangles

Chapter 16: Circles

Chapter 17: Constructions

Chapter 18: Surface Area and Volume of a Cuboid and a Cube

Chapter 19: Surface Area and Volume of a Right Circular Cylinder

Chapter 20: Surface Area and Volume of a Right Circular Cone

Chapter 21: Surface Area and Volume of a Sphere

Chapter 22: Tabular Representation of Statistical Data

Chapter 23: Graphical Representation of Statistical Data

Chapter 24: Measures of Central Tendency

Chapter 25: Probability

Chapter 1: Number Systems

Chapter 2: Exponents and Powers of Real Numbers

Chapter 3: Rationalisation

Chapter 4: Algebraic Identities

Chapter 5: Factorization of Algebraic Expressions

Chapter 6: Factorization of Polynomials

Chapter 7: Introduction to Euclidâ€™s Geometry

Chapter 8: Lines and Angles

Chapter 9: Triangle and Its Angles

Chapter 10: Congruent Triangles

Chapter 11: Coordinate Geometry

Chapter 12: Heronâ€™s Formula

Chapter 13: Linear Equations in Two Variables

Chapter 14: Quadrilaterals

Chapter 15: Areas of Parallelograms and Triangles

Chapter 16: Circles

Chapter 17: Constructions

Chapter 18: Surface Area and Volume of a Cuboid and a Cube

Chapter 19: Surface Area and Volume of a Right Circular Cylinder

Chapter 20: Surface Area and Volume of a Right Circular Cone

Chapter 21: Surface Area and Volume of a Sphere

Chapter 22: Tabular Representation of Statistical Data

Chapter 23: Graphical Representation of Statistical Data

Chapter 24: Measures of Central Tendency

Chapter 25: Probability

Class 9 RD Sharma Solutions – Chapter 2 Exponents of Real Numbers- Exercise 2.2 | Set 2

Question 12. Determine (8x)x, if 9x+2 = 240 + 9x.

Question 13. If 3x+1 = 9x−2, find the value of 21+x.

Question 14. If 34x = (81)−1 and (10)1/y = 0.0001, find the value of 2−x+4y.

Question 15. If 53x = 125 and 10y = 0.001. Find x and y.

Question 16. Solve the following equations:

(i) 3x+1 = 27 × 34

(ii)

(iii) 3x−1 × 52y−3 = 225

(iv) 8x+1 = 16y+2 and (1/2)3+x = (1/4)3y

(v) 4x−1 × (0.5)3−2x = (1/8)x

(vi)

Question: 17. If a and b are distinct positive primes such that, find x and y.

Question 18. If a and b are different positive primes such that,

(i) , find x and y.

(ii) (a + b)−1(a−1 + b−1) = axby, find x+y+2.

Question 19. If 2x × 3y × 5z = 2160, find x, y and z. Hence compute the value of 3x × 2−y × 5−z.

Question 20. If 1176 = 2a × 3b × 7c, find the values of a, b and c. Hence, compute the value of 2a × 3b × 7-c as a fraction.

Question 21. Simplify

(i)

(ii)

Question 22. Show that .

Question 23. (i) If a = xm+nyl, b = xn+lym and c = xl+myn, prove that am−n bn−l cl−m = 1.

(ii) If x = am+n, y = an+l and z = al+m, prove that xmynzl = xnylzm.

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Question 12. Determine (8x)^x, if 9^x+2 = 240 + 9^x.

Question 13. If 3^x+1 = 9^x−2, find the value of 2^1+x.

Question 14. If 3^4x= (81)⁻¹ and (10)^1/y = 0.0001, find the value of 2^−x+4y.

Question 15. If 5^3x = 125 and 10^y = 0.001. Find x and y.

(i) 3^x+1 = 27 × 3⁴

(ii) $4^{2x}=\left(\sqrt[3]{16}\right)^{\frac{-6}{y}}=(\sqrt{8})^2$

(iii) 3^x−1 × 5^2y−3 = 225

(iv) 8^x+1 = 16^y+2 and (1/2)^3+x = (1/4)^3y

(v) 4^x−1 × (0.5)^3−2x = (1/8)^x

(vi) $\sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2x}$

Question: 17. If a and b are distinct positive primes such that, $\sqrt[3]{a^6b^{-4}}=a^xb^{2y}$ find x and y.

(i) $\left(\frac{a^{-1}b^{2}}{a^2b^{-4}}\right)^7÷\frac{a^{3}b^{-5}}{a^{-2}b^{3}}=a^xb^y$ , find x and y.

(ii) (a + b)⁻¹(a⁻¹ + b⁻¹) = a^xb^y, find x+y+2.

Question 19. If 2^x × 3^y × 5^z = 2160, find x, y and z. Hence compute the value of 3^x × 2^−y × 5^−z.

Question 20. If 1176 = 2^a × 3^b × 7^c, find the values of a, b and c. Hence, compute the value of 2^a × 3^b × 7^-c as a fraction.

(i) $\left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c}\left(\frac{x^{c+a}}{x^b}\right)^{c-a}$

(ii) $\sqrt[lm]{\frac{x^l}{x^m}}×\sqrt[mn]{\frac{x^m}{x^n}}×\sqrt[nl]{\frac{x^n}{x^l}}$

Question 22. Show that $\frac{(a+\frac{1}{b})^m×(a-\frac{1}{b})^n}{(b+\frac{1}{a})^m×(b-\frac{1}{a})^n}=(\frac{a}{b})^{m+n}$ .

Question 23. (i) If a = x^m+ny^l, b = x^n+ly^m and c = x^l+myⁿ, prove that a^m−nb^n−lc^l−m = 1.

(ii) If x = a^m+n, y = a^n+land z = a^l+m, prove that x^myⁿz^l = xⁿy^lz^m.