RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

Read RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

Exercise 2.1
 
Question 1. Simplify the following:-
(i) 3(a4b3)10 × 5(a2b2)3
(ii) (2x -2 y3)3
(iii) ((4 × 107 )  (6 × 10-5))/(8 × 104
(iv) (4ab2  (-5ab3 ))/(10a2 b2 )
(v) {(x2 y2)/(a2 b2 )}n
(vi) (a3n-9)6/a2n-4
Solution 1. 

(i) We have 3(a4b3)10 × 5(a2b2)3

By using identity:-  (xm)n = xmn

After simplification we get

= 3(a40b30) × 5(a6b6)

= 3 × a40 × b30 × 5 × a6 × b6

By using identity:-  xm × xn = xm+n

= 15 × a46 × b36

= 15a46 b36                                                                  Answer.

 

(ii) (2x -2 y3)3

By using identity:- (x-m)= 1/x and (xm)n = xmn

After simplification we get

= 23 × 1/x2x3 × y3×3

= 8 × 1/x6 × y9

= 8  y9

= 8 x-6 y6                                                                  Answer.


RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number


Question 2. If a = 3 and b = - 2, find the value of:
(i) aa + bb
(ii) ab + ba
(iii) (a+b)ab
Solution 2.
(i) We have  aa + bb
Put the value a = 3 and b = -2 in equation given above
= 33 + (-2)-2 
= 3 × 3 × 3 + 1/((-22))
= 27 +1/((-2 ×-2))
= 27 +1/4
= 27/1+1/4
Take the LCM of denominators 1 and 4 = 4
= (108+1)/4
= 109/4                                                                        Answer.

(ii) We have  ab + ba
Put the value a = 3 and b = –2 in equation given above
= 3-2 + (–2)3
= 1/((32)) + (–2 × –2 × –2)   
= 1/9 – 8
= 1/9-8/1
Take the LCM of denominators 9 and 1 = 9
= (1-72)/9
= -71/9                                                                       Answer.
 
(iii) We have (a+b)ab
Put the value a = 3 and b = –2 in equation given above
= {3 + (–2)}3×(–2)
= {3 + (–2)}–6
= {3 – 2}–6
= (1) –6
= 1/16 
= 1                                                                              Answer.
 
 
Question 3.  Prove that:

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

Solution 3.

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number


 

Question 4. Prove that: 
 RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
 

 

Solution 4.
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number


Question 5. Prove that:
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
 
Solution 5.
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number


Question 8. Solve the following equation for x:
(i) 7(2x+3) = 1 
(ii) 2(x+1)= 4(x-3)
(iii) 2(5x+3) = 8(x+3)
(iv) 42x = 1/32
(v) 4(x-1)×(0.5)(3-2x) = (1/8)x
(vi) 2(3x-7) = 256
Solution 8. 
(i) we have 7(2x+3) = 1
By using identity:- x0 =1
7(2x+3) = 70
We should calculate the value of x through their powers.
2x + 3 = 0
2x = -3
x = (-3)/2                                                                             Answer.


(ii) we have 2(x+1)= 4(x-3)
By using identity:- (xm)n = xmn
2(x+1)= (22)(x-3) 
2(x+1)= (2)2(x-3)
2(x+1) = (2)(2x-6)
We should calculate the value of x through their powers.
x + 1 = 2x – 6 
x – 2x = – 6 – 1
– x = – 7 
x = 7                                                                             Answer.


(iii) we have 2(5x+3) = 8(x+3)
By using identity:- (xm)n = xmn
2(5x+3) = (23)(x+3)
2(5x+3)=(2)3(x+3)
2(5x+3)=2(3x+9) 
We should calculate the value of x through their powers.
5x+3 = 3x+9
5x – 3x = 9 – 3 
2x = 6
x = 6/2
x = 3                                                                             Answer.


(iv) we have 42x = 1/32
(4)2x = 1/(2)
(22)2x = 1/(2)
By using identity:- (xm)n = xmn
(2)4x = 2(-5)
We should calculate the value of x through their powers.
4x = -5 
x = -5/4                                                                           Answer.

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

(vi) 2(3x-7) = 256
We should calculate the value of x through their powers.
2(3x-7) = 28
3x - 7 = 8 
3x = 8 + 7
3x = 15
x = 15/3
x = 5                                                                          Answer.

 
Question 9.  Solve the following equations for x:
(i) 22x -22x+3+24 = 0
(ii) 3(2x+4)+1=2×3(x+2)
Solution 9. 
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
 
 
Question 10. If 49392 = a4b2c3, find the value of a, b and c where a, b and c are different positive primes.
Solution 10.
 
It is Given that 49392 = a4b2c3
By prime factorisation method we get
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
By prime factorisation we can observe that 24,32 and 73.
Hence the value of a = 2, b = 3 and c = 7                                                          Answer.
 
Question 11. If 1176 = 2a3b7c, find a,b and c.
Solution 11
It is Given that 1176 = 2a3b7c
By prime factorisation method we get
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
By prime factorisation we can observe that 23,31 and 72.
Hence the value of a = 3, b = 1 and c = 2                                                          Answer.
 
Question 12.  Given 4725 = 3a 5b 7c, find 
(i) the integral value of a, b and c
(ii) the value of 2-a3b7c
Solution 12.
(i) It is Given that 4725 = 3a 5b 7c
By prime factorisation method we get
 RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
By prime factorisation we can observe that 33,52 and 71.
Hence the value of a = 3, b = 2 and c = 1.
 
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
 
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
 

Exercise 2.2


Question 1.  Assuming that x,y,z are positive real numbers, simplify each of the following:
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

Solution 1.

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number


RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number


RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number


RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

 

Exercise 2.3  
Question 1. Write (625)-1/4 in decimal form. 
Solution  1. 
We have (625)-1/4  
By the prime factorisation 625 we get 5×5×5×5 
= (54)-1/4 
= 5-4/4   
= 5-1 
= 1/5  
= 0.2
 
Question 2.  State the product law of exponents. 
Solution  2.  
The product law of exponents is, “If a is any real number and m, n are positive integers, then x× x= x(m+n)
With the reference of above definition we get, 
x× x=(x×x×x……………to m factors)×(x×x×x……………to n factors)  
= a×a×a……………….. to (m+n) factors 
= a(m+n) 
Hence, am×an = a(m+n)
 
Question 3. State the quotient law of exponents.
Solution 3.
In the quotient law of exponents we need to divide two powers with the same base as given in example:- xm/xn 
Here a≠0.
When base is same and exponents in division we should subtracting the exponents. Here m and n are the positive integers, then xm/xn =x(m-n)
 
Question 4. State the power law of exponents.
Solution  4.
In the power rule of exponents we need to raise a power to a power by multiply the exponents with each other.  If a is any real number and m, n are positive integers, then
each other.  If a is any real number and m, n are positive integers, then
(xm )=xm× xm× x× xm……………..n factors
(xm )=(x×x×x×…..m)× (x×x×x×…..n) 
(xm )=(x×x×x…..mn) factors
(xm )= xmn 
 

Question 5. If 24×42=16x, then find the value of x.

Solution 5.

We have 24×4= 16x

By the prime factorisation of 16 we get 2×2×2×2.

24×42  = (24 )x 

24×(22)2=(24)

2× 24=(24)x 

28=24x 

By equating the exponents we get

8 = 4x

8/4=x 

2=x                                                                               Answer


Question 6. If 3x-1=9 and 4y+2=64, what is the value of x/y?
Solution 6. 
We have 3x-1=9
By the prime factorisation of 9 we get 3×3
3x-1 = 3×3 
3x-1 = 32 
By equating the exponents we get
x-1=2 
x=2+1 
x=3 
 
4y+2 = 64 
By the prime factorisation of 64 we get 4×4×4
4y+2 = 4×4×4 
4y+2 = 43 
By equating the exponents we get
y+2=3 
y=3-2 
y=1 
Therefore,
x/y=3/1 
x/y=3                                                                              Answer

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number


RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number




Question 14. If (x-1)3= 8, What is the value of (x+1)?
Solution 14.
We have (x-1)= 8
By the prime factorisation of 8 we get 2×2×2
(x-1)= 23 
Equalising the base we get
x-1=2 
x=2+1 
x=3 
Value of x put in given equation (x+1)2
=(3+1)
=(4)2   
=4×4 
=16                                                                              Answer
 
NCERT Exemplar Solutions Class 9 Maths Areas of Parallelogram and Triangle
NCERT Exemplar Solutions Class 9 Maths Circle
NCERT Exemplar Solutions Class 9 Maths Constructions
NCERT Exemplar Solutions Class 9 Maths Coordinate Geometry
NCERT Exemplar Solutions Class 9 Maths Euclids Geometry
NCERT Exemplar Solutions Class 9 Maths Herons Formula
NCERT Exemplar Solutions Class 9 Maths Linear Equations in two variables
NCERT Exemplar Solutions Class 9 Maths Lines and Angles
NCERT Exemplar Solutions Class 9 Maths Number System
NCERT Exemplar Solutions Class 9 Maths Polynomials
NCERT Exemplar Solutions Class 9 Maths Probability
NCERT Exemplar Solutions Class 9 Maths Quadrilaterals
NCERT Exemplar Solutions Class 9 Maths Statistics
NCERT Exemplar Solutions Class 9 Maths Surface areas and Volumes
NCERT Exemplar Solutions Class 9 Maths Triangles
RS Aggarwal Class 9 Mathematics Solutions Chapter 1 Real Numbers
RS Aggarwal Class 9 Mathematics Solutions Chapter 2 Polynomials
RS Aggarwal Class 9 Mathematics Solutions Chapter 3 Introduction to Euclids Geometry
RS Aggarwal Class 9 Mathematics Solutions Chapter 4 Lines and Triangles
RS Aggarwal Class 9 Mathematics Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle
RS Aggarwal Class 9 Mathematics Solutions Chapter 6 Coordinate Geometry
RS Aggarwal Class 9 Mathematics Solutions Chapter 7 Areas by herons formula
RS Aggarwal Class 9 Mathematics Solutions Chapter 8 Linear Equations in Two Variables
RS Aggarwal Class 9 Mathematics Solutions Chapter 9 Quadrilaterals and Parallelograms
RS Aggarwal Class 9 Mathematics Solutions Chapter 10 Area
RS Aggarwal Class 9 Mathematics Solutions Chapter 11 Circle
RS Aggarwal Class 9 Mathematics Solutions Chapter 12 Geometrical Constructions
RS Aggarwal Class 9 Mathematics Solutions Chapter 13 Volume and Surface Area
RS Aggarwal Class 9 Mathematics Solutions Chapter 14 Statistics
RS Aggarwal Class 9 Mathematics Solutions Chapter 15 Probability
RD Sharma Solutions Class 9 Maths
RD Sharma Solutions Class 9 Chapter 1 Number System
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 3 Rationalisation
RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
RD Sharma Solutions Class 9 Chapter 5 Factorization of Algebraic Expressions
RD Sharma Solutions Class 9 Chapter 6 Factorization of Polynomials
RD Sharma Solutions Class 9 Chapter 7 Introduction To Euclids Geometry
RD Sharma Solutions Class 9 Chapter 8 Lines and Angles
RD Sharma Solutions Class 9 Chapter 9 Triangle and its Angles
RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles
RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
RD Sharma Solutions Class 9 Chapter 15 Area of Parallelograms and Triangles
RD Sharma Solutions Class 9 Chapter 16 Circles
RD Sharma Solutions Class 9 Chapter 17 Construction
RD Sharma Solutions Class 9 Chapter 18 Surface Area and Volume of Cuboid and Cube
RD Sharma Solutions Class 9 Chapter 19 Surface Area and Volume of A Right Circular Cylinder
RD Sharma Solutions Class 9 Chapter 20 Surface Area and Volume of A Right Circular Cone
RD Sharma Solutions Class 9 Chapter 21 Surface Area and Volume of A Sphere
RD Sharma Solutions Class 9 Chapter 22 Tabular Representation of Statistical Data
RD Sharma Solutions Class 9 Chapter 23 Graphical Representation of Statistical Data
RD Sharma Solutions Class 9 Chapter 24 Measure of Central Tendency
RD Sharma Solutions Class 9 Chapter 25 Probability

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