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

Solution:

We have,

Solution:

We have,

Solution:

We have,

Solution:

We have,

Solution:

We have,

= 3 x2 y z2

Solution:

We have,

Solution:

We have,

Solution:

We have,

Solution:

We have,

= 2âˆ’3

Solution:

We have,

= 7âˆ’2

Solution:

We have,

= 0.1

Solution:

We have,

Solution:

We have,

Solution:

We have,

= 52 Ã— 7

= 175

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 27 âˆ’ 3 âˆ’ 9

= 15

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 2 Ã— 1 Ã— 5

= 10

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 1 +

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 28âˆš2

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

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.

Solution:

We have,

L.H.S. =

= x0

= 1

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

### (v) (xa-b)a+b (xb-c)b+c (xc-a)c+a = 1

Solution:

We have,

L.H.S. = (xa-b)a+b (xb-c)b+c (xc-a)c+a

= x0

= 1

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= x

= R.H.S.

Hence proved.

### (vii)

Solution:

We have,

L.H.S. =

= (ax-y)x+y (ay-z)y+z (ax-z)x+z

= a0

= 1

= R.H.S.

Hence proved.

### (viii)

Solution:

We have,

L.H.S. =

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

= 30

= 1

= R.H.S.

Hence proved.

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

Solution:

We are given,

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

So, we get,

=> 12 = k1/z

=> 2 Ã— 3 Ã— 2 = k1/z

=> 22 Ã— 3 = k1/z

=> (k1/x)2 Ã— (k)1/y = k1/z

=> (k)2/x Ã— (k)1/y = k1/z

=>  = k1/z

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

Hence proved.

### Question 6. If 2x = 3y = 6âˆ’z, show that 1/x + 1/y + 1/z = 0.

Solution:

We are given,

=> 2x = 3y = 6âˆ’z = k (say)

So, we get,

=> 6 = k-1/z

=> (2 Ã— 3) = k-1/z

=> k1/x Ã— k1/y = k-1/z

=>  = k-1/z

=> 1/x + 1/y = âˆ’1/z

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

Hence proved.

### Question 7. If ax = by = cz and b2 = ac, then show that y = 2zx/(z+x).

Solution:

We are given,

=> ax = by = cz = k (say)

=> a = k1/x, b = k1/y, c = k1/z

We are given, b2 = ac

=> (k1/y)2 = k1/x Ã— k1/z

=> k2/y

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

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

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

Hence proved.

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

Solution:

We are given,

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

So, we get,

=> 75 = k1/z

=> 3 Ã— 52 = k1/z

=> (k)1/x Ã— (k1/y)2 = k1/z

=> (k)1/x Ã— (k)2/y = k1/z

=>  = k1/z

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

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

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

Hence proved.

Solution:

We are given,

=> (27)x = 9/3x

=> (33)x = 32/3x

=> 33x = 32âˆ’x

=> 3x = 2 âˆ’ x

=> 4x = 2

=> x = 2/4

=> x = 1/2

Solution:

We have,

=> 25x Ã· 2x

=> 25xâˆ’x

=> 24x = 24

=> 4x = 4

=> x = 1

Solution:

We have,

=> (23)4 = (22)x

=> 212 = 22x

=> 2x = 12

=> x = 6

Solution:

We have,

=>

=>

=>

=>

=>

=> x = 3

### (iv) 5xâˆ’2 Ã— 32xâˆ’3 = 135

Solution:

We have,

=> 5xâˆ’2 Ã— 32xâˆ’3 = 135

=> 5xâˆ’2 Ã— 32xâˆ’3 = 5 Ã— 27

=> 5xâˆ’2 Ã— 32xâˆ’3 = 51 Ã— 33

=> x âˆ’ 2 = 1 and 2x âˆ’ 3 = 3

=> x = 3

### (v) 2xâˆ’7 Ã— 5xâˆ’4 = 1250

Solution:

We are given,

=> 2xâˆ’7 Ã— 5xâˆ’4 = 1250

=>  2xâˆ’7 Ã— 5xâˆ’4 = 2 Ã— 625

=> 2xâˆ’7 Ã— 5xâˆ’4 = 2 Ã— 54

=> x âˆ’ 7 = 1 and x âˆ’ 4 = 4

=> x = 8

### (vi)

Solution:

We have,

=>

=>

=>

=> 4x/3 + 1/3 = âˆ’5

=> 4x +1 = âˆ’15

=> 4x = âˆ’16

=> x = âˆ’4

Solution:

We have,

=> 52x+3 = 1

=> 52x+3 = 50

=> 2x + 3 = 0

=> 2x = âˆ’3

=> x = âˆ’3/2

### (viii)

Solution:

We have,

=>

=>  = 256 âˆ’ 81 âˆ’ 6

=>  = 169

=>

=> âˆšx = 2

=> x = 4

### (ix)

Solution:

We have,

=>

=>

=>

=> (x+1)/2 = âˆ’3

=> x + 1 = âˆ’6

=> x = âˆ’7

### Question 11. If x = 21/3 + 22/3, show that x3 âˆ’ 6x = 6.

Solution:

Given, x = 21/3 + 22/3

Therefore, x3 = (21/3)3 + (22/3)3 + 3(21/3)(22/3)(21/3 + 22/3)

=> x3 = (21/3)3 + (22/3)3 + 3(21/3)(22/3)(x)

=> x3 = 2 + 4 + 3(2)(x)

=> x3 = 6 + 6x

=> x3 âˆ’ 6x = 6

Hence proved.

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