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

Set 2 of Exercise 2.2 in Chapter 2 typically delves deeper into the application of exponent laws. This set often introduces more complex problems that require students to combine multiple laws of exponents, work with negative and fractional exponents, and simplify more intricate expressions. The questions in this set are designed to challenge students' understanding and help them develop problem-solving skills in the context of exponents.

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)  

Solution:

We have,

=> 

=> 

=> 

=> 

=> 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)  

Solution:

We have,

=> 

=> 

=> 1/2 = 2x − 1

=> 2x = 3/2

=> x = 3/4

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

Solution:

We have,

=> 

=> (a⁶ b⁻⁴)^1/3 = a^xb^2y

=> a^6/3 b^−4/3 = a^xb^2y

=> a² b^−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) , find x and y.

Solution:

We have,

=> 

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

=> (a⁻³ b⁶)⁷ ÷ (a⁵ b⁻⁸) = a^xb^y

=> (a⁻²¹ b⁴²) ÷ (a⁵ b⁻⁸) = a^xb^y

=> (a⁻²¹⁻⁵ b⁴²⁺⁸) = a^xb^y

=> (a⁻²⁶ b⁵⁰) = 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

=>  = a^xb^y

=>  = 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) 

Solution:

We have,

= 

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

= 

= 

= x⁰

= 1

(ii) 

Solution:

We have,

=> 

=> 

=> 

=> 

=> 

=> 

=> x⁰

= 1

Question 22. Show that .

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 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−n b^n−l c^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

= 

= 

= x⁰y⁰

= 1

= R.H.S.

Hence proved.

(ii) If x = a^m+n, y = a^n+l and 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+n)ⁿ (a^n+l)^l (a^l+m)^m

= xⁿy^lz^m

= R.H.S.

Hence proved.

Summary

Set 2 of Exercise 2.2 builds upon the foundational knowledge of exponent laws introduced in earlier sections. It challenges students to apply these laws in more complex scenarios, often requiring the combination of multiple rules. This set typically includes problems involving negative and fractional exponents, which helps students understand the full range of exponent applications. The questions are designed to improve critical thinking and problem-solving skills, preparing students for more advanced mathematical concepts they will encounter in future studies.