RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number

(i) We have 3(a⁴b³)¹⁰ × 5(a²b²)³

By using identity:-  (x^m)ⁿ = x^mn

After simplification we get

= 3(a⁴⁰b³⁰) × 5(a⁶b⁶)

= 3 × a⁴⁰ × b³⁰ × 5 × a⁶ × b⁶

By using identity:-  x^m × xⁿ = x^m+n

= 15 × a⁴⁶ × b³⁶

= 15a⁴⁶ b³⁶                                                                  Answer.

(ii) (2x ^-2 y³)³

By using identity:- (x^-m)= 1/x^m  and (x^m)ⁿ = x^mn

After simplification we get

= 2³ × ¹/x^2x3 × y^3×3

= 8 × 1/x⁶ × y⁹

= 8  y⁹

= 8 x^-6 y⁶                                                                  Answer.

Solution 3.

 

Question 5. Prove that:

 

Solution 5.

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) 4^2x = 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:- (x^m)ⁿ = x^mn

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:- (x^m)ⁿ = x^mn

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 4^2x = 1/32

(4)^2x = 1/(2)⁵ 

(2²)^2x = 1/(2)⁵ 

By using identity:- (x^m)ⁿ = x^mn

(2)^4x = 2^(-5)

We should calculate the value of x through their powers.

4x = -5 

x = -5/4                                                                           Answer.

(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) 2^2x -2^2x+3+2⁴ = 0

(ii) 3^(2x+4)+1=2×3^(x+2)

Solution 9. 

 

 

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

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 = 2^a3^b7^c, find a,b and c.

Solution 11

It is Given that 1176 = 2^a3^b7^c

By prime factorisation method we get

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 = 3^a 5^b 7^c, find 

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

(ii) the value of 2-a3b7c

Solution 12.

(i) It is Given that 4725 = 3^a 5^b 7^c

By prime factorisation method we get

 

By prime factorisation we can observe that 3³,5² and 7¹.

Hence the value of a = 3, b = 2 and c = 1.

 

Exercise 2.2

Question 1.  Assuming that x,y,z are positive real numbers, simplify each of the following:

Solution 1.

 

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^m × xⁿ = x^(m+n)

With the reference of above definition we get, 

x^m × xⁿ =(x×x×x……………to m factors)×(x×x×x……………to n factors)  

= a×a×a……………….. to (m+n) factors 

= a^(m+n) 

Hence, a^m×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:- x^m/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 x^m/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

(x^m )ⁿ =x^m× x^m× x^m × x^m……………..n factors

(x^m )ⁿ =(x×x×x×…..m)× (x×x×x×…..n) 

(x^m )ⁿ =(x×x×x…..mn) factors

(x^m )ⁿ = x^mn 

 

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

Solution 5.

We have 2⁴×4² = 16^x

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

2⁴×4²  = (2⁴ )^x 

2⁴×(2²)²=(2⁴)^x 

2⁴ × 2⁴=(2⁴)^x 

2⁸=2^4x 

By equating the exponents we get

8 = 4x

8/4=x 

2=x                                                                               Answer

Question 6. If 3^x-1=9 and 4^y+2=64, what is the value of x/y?

Solution 6. 

We have 3^x-1=9

By the prime factorisation of 9 we get 3×3

3^x-1 = 3×3 

3^x-1 = 3² 

By equating the exponents we get

x-1=2 

x=2+1 

x=3 

 

4^y+2 = 64 

By the prime factorisation of 64 we get 4×4×4

4^y+2 = 4×4×4 

4^y+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

Question 14. If (x-1)³= 8, What is the value of (x+1)² ?

Solution 14.

We have (x-1)³ = 8

By the prime factorisation of 8 we get 2×2×2

(x-1)³ = 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)² 

=(4)²   

=4×4 

=16                                                                              Answer