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

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

(i) 

Solution:

We have,

= 

= 

= 

= 

(ii) 

Solution:

We have,

= 

= 

= 

(iii) 

Solution:

We have,

= 

= 

= 

(iv) 

Solution:

We have,

= 

= 

= 

= 

= 

(v) 

Solution:

We have,

= 

= 

= 

= 3 x² y z²

(vi) 

Solution:

We have,

= 

= 

= 

= 

(vii) 

Solution:

We have,

= 

= 

= 

= 

Question 2. Simplify

(i) 

Solution:

We have,

= 

= 

= 

= 

(ii) 

Solution:

We have,

= 

= 

= 2⁻³

= 

(iii) 

Solution:

We have,

= 

= 

= 7⁻²

= 

(iv) 

Solution:

We have,

= 

= 

= 

= 0.1

(v) 

Solution:

We have,

= 

= 

= 

= 

= 

(vi) 

Solution:

We have,

= 

= 

= 

= 

= 

(vii) 

Solution:

We have,

= 

= 

= 

= 5² × 7

= 175

Question 3. Prove that

(i) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= 

= 

= R.H.S.

Hence proved.

(ii) 

Solution:

We have,

L.H.S. = 

= 

= 

= 27 − 3 − 9

= 15

= R.H.S.

Hence proved.

(iii) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= R.H.S.

Hence proved.

(iv) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= 2 × 1 × 5

= 10

= R.H.S.

Hence proved.

(v) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= R.H.S.

Hence proved.

(vi) 

Solution:

We have,

L.H.S. = 

= 

= 1 + 

= 

= R.H.S.

Hence proved. 

(vii) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= R.H.S.

Hence proved. 

(viii) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= 28√2

= R.H.S.

Hence proved. 

(ix) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= R.H.S.

Hence proved. 

Question 4. Show that

(i) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 1

= R.H.S.

Hence proved. 

(ii) 

Solution:

We have,

L.H.S. = 

= 

= 

= [x^-2ab-(-2ab)]^a+b

= [x⁰]^a+b

= x⁰

= 1

= R.H.S.

Hence proved. 

(iii) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= x⁰

= 1

= R.H.S.

Hence proved. 

(iv) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= R.H.S.

Hence proved. 

(v) (x^a-b)^a+b (x^b-c)^b+c (x^c-a)^c+a = 1

Solution:

We have,

L.H.S. = (x^a-b)^a+b (x^b-c)^b+c (x^c-a)c+a 

= 

= 

= x⁰

= 1

= R.H.S.

Hence proved. 

(vi) 

Solution:

We have,

L.H.S. = 

= 

= 

= x

= R.H.S.

Hence proved. 

(vii) 

Solution:

We have,

L.H.S. = 

= (a^x-y)^x+y (a^y-z)^y+z (a^x-z)^x+z 

= 

= 

= a⁰

= 1

= R.H.S.

Hence proved. 

(viii) 

Solution:

We have,

L.H.S. = 

= (3^a-b)^a+b (3^b-c)^b+c (3^c-a)^c+a 

= 

= 

= 3⁰

= 1

= R.H.S.

Hence proved. 

Question 5. If 2^x = 3^y = 12^z, show that 1/z = 1/y + 2/x.

Solution:

We are given,

=> 2^x = 3^y = 12^z = k (say)

So, we get,

=> 12 = k^1/z

=> 2 × 3 × 2 = k^1/z

=> 2² × 3 = k^1/z

=> (k^1/x)² × (k)^1/y = k^1/z

=> (k)^2/x × (k)^1/y = k^1/z

=>  = k^1/z

=> 2/x + 1/y = 1/z

Hence proved.

Question 6. If 2^x = 3^y = 6^−z, show that 1/x + 1/y + 1/z = 0.

Solution:

We are given,

=> 2^x = 3^y = 6^−z = k (say)

So, we get,

=> 6 = k^-1/z

=> (2 × 3) = k^-1/z

=> k^1/x × k^1/y = k^-1/z

=>  = k^-1/z

=> 1/x + 1/y = −1/z

=> 1/x + 1/y + 1/z = 0

Hence proved.

Question 7. If a^x = b^y = c^z and b² = ac, then show that y = 2zx/(z+x).

Solution:

We are given,

=> a^x = b^y = c^z = k (say)

=> a = k^1/x, b = k^1/y, c = k^1/z

We are given, b² = ac

=> (k^1/y)² = k^1/x × k^1/z

=> k^2/y = 

=> 2/y = 1/x + 1/z

=> 2/y = (x+z)/xz

=> y = 2zx/(z+x)

Hence proved.

Question 8. If 3^x = 5^y = (75)^z, show that z = xy/(2x+y).

Solution:

We are given,

=> 3^x = 5^y = (75)^z = k (say)

So, we get,

=> 75 = k^1/z

=> 3 × 5² = k^1/z

=> (k)^1/x × (k^1/y)² = k^1/z

=> (k)^1/x × (k)^2/y = k^1/z

=>  = k^1/z

=> 1/x + 2/y = 1/z

=> (2x+y)/xy = 1/z

=> z = xy/(2x+y)

Hence proved. 

Question 9. If (27)^x = 9/3^x, find x.

Solution:

We are given,

=> (27)^x = 9/3^x

=> (3³)^x = 3²/3^x

=> 3^3x = 3^2−x

=> 3x = 2 − x

=> 4x = 2

=> x = 2/4

=> x = 1/2

Question 10. Find the values of x in each of the following:

(i) 2^5x ÷ 2^x =

Solution:

We have,

=> 2^5x ÷ 2^x = 

=> 2^5x−x = 

=> 2^4x = 2⁴

=> 4x = 4

=> x = 1

(ii) (2³)⁴ = (2²)^x

Solution:

We have,

=> (2³)⁴ = (2²)^x

=> 2¹² = 2^2x

=> 2x = 12

=> x = 6

(iii) 

Solution:

We have,

=> 

=> 

=> 

=> 

=> 

=> x = 3

(iv) 5^x−2 × 3^2x−3 = 135

Solution:

We have,

=> 5^x−2 × 3^2x−3 = 135

=> 5^x−2 × 3^2x−3 = 5 × 27

=> 5^x−2 × 3^2x−3 = 5¹ × 3³

=> x − 2 = 1 and 2x − 3 = 3 

=> x = 3

(v) 2^x−7 × 5^x−4 = 1250

Solution:

We are given,

=> 2^x−7 × 5^x−4 = 1250

=>  2^x−7 × 5^x−4 = 2 × 625

=> 2^x−7 × 5^x−4 = 2 × 5⁴

=> x − 7 = 1 and x − 4 = 4 

=> x = 8

(vi)

Solution:

We have,

=> 

=> 

=> 

=> 4x/3 + 1/3 = −5

=> 4x +1 = −15

=> 4x = −16

=> x = −4

(vii) 5^2x+3 = 1

Solution:

We have,

=> 5^2x+3 = 1

=> 5^2x+3 = 5⁰

=> 2x + 3 = 0

=> 2x = −3

=> x = −3/2

(viii) 

Solution:

 We have,

=> 

=>  = 256 − 81 − 6

=>  = 169 

=> 

=> √x = 2

=> x = 4

(ix)

Solution:

We have,

=> 

=> 

=> 

=> (x+1)/2 = −3

=> x + 1 = −6

=> x = −7

Question 11. If x = 2^1/3 + 2^2/3, show that x³ − 6x = 6.

Solution:

Given, x = 2^1/3 + 2^2/3

Therefore, x³ = (2^1/3)³ + (2^2/3)³ + 3(2^1/3)(2^2/3)(2^1/3 + 2^2/3)

=> x³ = (2^1/3)3 + (2^2/3)³ + 3(2^1/3)(2^2/3)(x)

=> x³ = 2 + 4 + 3(2)(x)

=> x³ = 6 + 6x

=> x³ − 6x = 6

Hence proved.