RD Sharma Solutions Ex-2.1, Exponents Of Real Numbers, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q.1. Simplify the following: 

(i) 3(a⁴ b³)¹⁰ x 5 (a² b²)³ 

Solution: = 3(a⁴⁰ b³⁰) x 5 (a⁶ b⁶) = 15 (a⁴⁶ b³⁶) 

(ii) (2x ^-2 y³)³

Solution:  = (2³ x ^-2x3 y^3x3) = 8x ^-6 y⁹

(iii) 

Solution: 

(iv) 

Solution: = 

(v)

Solution: 

(vi) 

Solution:    = a¹⁶ⁿ⁻⁵⁰ 

2. If a = 3 and b = -2, find the values of: 

(i) a^a+b^b 
(ii) a^b+b^a 
(iii) a^b+b^a 

Solution: (i) We have,a^a+b^b = 3³+(−2)⁻² = 3³+  = 27+1/4 = 109/4 

(ii) a^b+b^a = 3⁻²+(−2)³ = (1/3)²+(−2)³

(iii) We have, a^b+b^a = (3+(−2))³⁽⁻²⁾ = (3−2))⁻⁶ = 1⁻⁶ = 1 

3.Prove that: (i) 

(ii) 
(iii) 

Solution:   = 1

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Or, Therefore, LHS = RHS Hence proved

(ii) To prove, 

 Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved (iii) To prove,  

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved 

4. Prove that: (i) 

(ii)

Solution: (i)  

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved

Left hand side  (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved 

5.Prove that: (i) 

(ii) 

Solution: (i) To prove,   = abc

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

abc

Therefore, LHS = RHS Hence proved (ii) To prove, 

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved 

6. If abc = 1, show that 

Solution: To prove,

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

We know abc = 1

By substituting the value c in equation (1), we get

Therefore, LHS = RHS Hence proved 

7. Simplify: (i) 

Solution: 

(ii)  

Solution: = 

(iii) 

Solution: = 

(iv) 

Solution:

8. Solve the following equations for x: 

(i) 7^2x+3 = 1 
(ii) 2^x+1=4^x−3 
(iii) 2^5x+3 = 8^x+3 
(iv) 4^2x = 13/2 
(v) 4^x−1×(0.5)^3−2x=(1/8)^x 
(vi) 2^3x−7 = 256 

Solution: (i) We have,

⇒ 7^2x+3 = 1 ⇒ 7^2x+3 = 7⁰ ⇒ 2x + 3 = 0 ⇒ 2x = -3 ⇒ x

 (ii) We have,

(iii) We have,

(iv) We have,

(v) We have,

9. Solve the following equations for x:  

(i) 2^2x−2^x+3+2⁴ = 0 

(ii) 3^2x+4+1=2×3^x+2

Solution: 

(i) We have, ⇒ 2^2x−2^x+3+2⁴ = 0 

⇒  2^2x+2⁴=2^x.2³ 
⇒ Let ^2x=y 
⇒ y²+2⁴=y×2³ 
⇒  y²−8y+16=0 
⇒ y²−4y−4y+16=0 
⇒ y(y-4) -4(y-4) = 0
⇒ y = 4
⇒ x²=2² 
⇒ x = 2

(ii) We have,

3^2x+4+1=2×3^x+2

(3^x+2)²+1=2×3^x+2

Let 3^x+2 = y

y²+1=2y

y²−2y+1=0

y²−y−y+1=0

y(y−1)−1(y−1)=0

(y−1)(y−1)=0

y = 1

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

Solution: Taking out the LCM , the factors are 2⁴,3² and 7³ a⁴b²c³ = 2⁴,3² and 7³

a = 2, b = 3 and c = 7 [Since, a, b and c are primes]

11. If 1176 = 2^a×3^b×7^c, Find a, b, and c. 

Solution: Given that 2, 3 and 7 are factors of 1176. Taking out the LCM of 1176, we get 2³×3¹×7² = 2^a×3^b×7^c By comparing, we get

a = 3, b = 1 and c = 2.

12. Given 4725 = 3^a×5^b×7^c, find (i) The integral values of a, b and c 

Solution: Taking out the LCM of 4725, we get

3³×5²×7¹ = 3^a×5^b×7^c

By comparing, we get

a = 3, b = 2 and c = 1.

(ii) The value of 2^−a×3^b×7^c 

Solution:

 = 2⁻³×3²×7¹

2⁻³×3²×7¹ = 1/8×9×7

63/8

13. If a = xy^p−1, b = xy^q−1 and c = xy^r−1, prove that a^q−rb^r−pc^p−q = 1 

Solution: Given, a = xy^p−1, b = xy^q−1 and c = xy^r−1 

To prove, a^q−rb^r−pc^p−q = 1

Left hand side (LHS) = Right hand side (RHS) Considering LHS, = a^q−rb^r−pc^p−q        ……(i)

By substituting the value of a, b and c in equation

(i), we get

= (xy^p−1)^q−r(xy^q−1)r−p(xy^r−1)^p−q 

= xy^{pq−pr−q+r}xy^{qr−pq−r+p}xy^{rp−rq−p+q} 

= xy^{pq−pr−q+r+qr−pq−r+p+rp−rq−p+q} 

= xy⁰ = 1