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

• Last Updated : 30 Apr, 2021

### Question 12. Determine (8x)x, if 9x+2 = 240 + 9x.

Solution:

We have,

=> 9x+2 = 240 + 9x

=> 9x+2 âˆ’ 9x = 240

=> 9x (92 âˆ’ 1) = 240

=> 9x = 240/80

=> 32x = 3

=> 2x = 1

=> x = 1/2

Therefore, (8x)x = [8 Ã— (1/2)]1/2

= 41/2

= 2

### Question 13. If 3x+1 = 9xâˆ’2, find the value of 21+x.

Solution:

We have,

=> 3x+1 = 9xâˆ’2

=> 3x+1 = (32)xâˆ’2

=> 3x+1 = 32xâˆ’4

=> x + 1 = 2x âˆ’ 4

=> x = 5

Therefore, 21+x = 21+5

= 26

= 64

### Question 14. If 34x = (81)âˆ’1 and (10)1/y = 0.0001, find the value of 2âˆ’x+4y.

Solution:

We are given,

=> 34x = (81)âˆ’1

=> 34x = (34)âˆ’1

=> 34x = (3)âˆ’4

=> 4x = âˆ’4

=> x = âˆ’1

And also, (10)1/y = 0.0001

=> (10)1/y = (10)âˆ’4

=> 1/y = âˆ’4

=> y = âˆ’1/4

Therefore, 2âˆ’x+4y = 21+4(âˆ’1/4)

= 21âˆ’1

= 1

### Question 15. If 53x = 125 and 10y = 0.001. Find x and y.

Solution:

We are given,

=> 53x = 125

=> 53x = 53

=> 3x = 3

=> x =1

Also, (10)y = 0.001

=> 10y = 10âˆ’3

=> y = âˆ’3

Therefore, the value of x is 1 and the value of y is â€“3.

### (i) 3x+1 = 27 Ã— 34

Solution:

We have,

=>  3x+1 = 27 Ã— 34

=> 3x+1 = 33 Ã— 34

=> 3x+1 = 37

=> x + 1 = 7

=> x = 6

### (ii)

Solution:

We have,

=>

=>

=>

=>

=> 4x = âˆ’8/y = 3

=> x = 3/4 and y = âˆ’8/3

### (iii) 3xâˆ’1 Ã— 52yâˆ’3 = 225

Solution:

We have,

=> 3xâˆ’1 Ã— 52yâˆ’3 = 225

=> 3xâˆ’1 Ã— 52yâˆ’3 = 32 Ã— 52

=> x âˆ’ 1 = 2 and 2y âˆ’ 3 = 2

=> x = 3 and 2y = 5

=> x = 3 and y = 5/2

### (iv) 8x+1 = 16y+2 and (1/2)3+x = (1/4)3y

Solution:

We have,

=> 8x+1 = 16y+2

=> (23)x+1 = (24)y+2

=> 23x+3 = 24y+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 âˆ’ 3   . . . . (2)

Putting (2) in (1), we get,

=> 3(6y âˆ’ 3) + 3 = 4y + 8

=> 18y âˆ’ 9 + 3 = 4y + 8

=> 14y = 14

=> y = 1

Putting y = 1 in (2), we get,

x = 6(1) âˆ’ 3 = 6 âˆ’ 3 = 3

Therefore, the value of x is 1 and the value of y is â€“3.

### (v) 4xâˆ’1 Ã— (0.5)3âˆ’2x = (1/8)x

Solution:

We have,

=> 4xâˆ’1 Ã— (0.5)3âˆ’2x = (1/8)x

=> (22)xâˆ’1 Ã— (1/2)3âˆ’2x = [(1/2)3]x

=> 22xâˆ’2 Ã— 22xâˆ’3 = 2âˆ’3x

=> 22xâˆ’2+2xâˆ’3 = 2âˆ’3x

=> 24xâˆ’5 = 2âˆ’3x

=> 4x âˆ’ 5 = âˆ’3x

=> 7x = 5

=> x = 5/7

### (vi)

Solution:

We have,

=>

=>

=> 1/2 = 2x âˆ’ 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 bâˆ’4)1/3 = axb2y

=> a6/3 bâˆ’4/3 = axb2y

=> a2 bâˆ’4/3 = axb2y

=> x = 2 and 2y = âˆ’4/3

=> x = 2 and y = âˆ’2/3

### (i) , find x and y.

Solution:

We have,

=>

=> (aâˆ’1âˆ’2 b2+4)7 Ã· (a3+2 bâˆ’5âˆ’3) = axby

=> (aâˆ’3 b6)7 Ã· (a5 bâˆ’8) = axby

=> (aâˆ’21 b42) Ã· (a5 bâˆ’8) = axby

=> (aâˆ’21âˆ’5 b42+8) = axby

=> (aâˆ’26 b50) = axby

=> x = âˆ’26, y = 50

### (ii) (a + b)âˆ’1(aâˆ’1 + bâˆ’1) = axby, find x+y+2.

Solution:

We have,

=> (a + b)âˆ’1(aâˆ’1 + bâˆ’1) = axby

=>  = axby

=>  = axby

=> 1/ab = axby

=> aâˆ’1bâˆ’1 = axby

=> x = âˆ’1 and y = âˆ’1

So, x+y+2 = âˆ’1âˆ’1+2 = 0.

### Question 19. If 2x Ã— 3y Ã— 5z = 2160, find x, y and z. Hence compute the value of 3x Ã— 2âˆ’y Ã— 5âˆ’z.

Solution:

We are given,

=> 2x Ã— 3y Ã— 5z = 2160

=> 2x Ã— 3y Ã— 5z = 24 Ã— 33 Ã— 51

=> x = 4, y = 3, z = 1

Therefore, 3x Ã— 2âˆ’y Ã— 5âˆ’z = 34 Ã— 2âˆ’3 Ã— 5âˆ’1

= (81) (1/8) (1/5)

= 81/40

### Question 20. If 1176 = 2a Ã— 3b Ã— 7c, find the values of a, b and c. Hence, compute the value of 2a Ã— 3b Ã— 7-c as a fraction.

Solution:

We are given,

=> 1176 = 2a Ã— 3b Ã— 7c

=> 23 Ã— 31 Ã— 72 = 2a Ã— 3b Ã— 7c

=> a = 3, b = 1, c = 2

Therefore, 2a Ã— 3b Ã— 7âˆ’c = 23 Ã— 31 Ã— 7âˆ’2

= (8) (3) (1/49)

= 24/49

### (i)

Solution:

We have,

= (xa+bâˆ’c)aâˆ’b (xb+câˆ’a)bâˆ’c (xc+aâˆ’b)câˆ’a

= x0

= 1

Solution:

We have,

=>

=>

=>

=>

=>

=>

=> x0

= 1

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

### Question 23. (i) If a = xm+nyl, b = xn+lym and c = xl+myn, prove that amâˆ’n bnâˆ’l clâˆ’m = 1.

Solution:

Given, a = xm+nyl, b = xn+lym and c = xl+myn

We have,

L.H.S. = amâˆ’n bnâˆ’l clâˆ’m

= (xm+nyl)mâˆ’n(xn+lym)nâˆ’l(xl+myn)lâˆ’m

= x0y0

= 1

= R.H.S.

Hence proved.

### (ii) If x = am+n, y = an+l and z = al+m, prove that xmynzl = xnylzm.

Solution:

Given, x = am+n, y = an+l and z = al+m.

We have,

L.H.S. = xmynzl

= (am+n)m (an+l)n (al+m)l

= (am+n)n (an+l)l (al+m)m

= xnylzm

= R.H.S.

Hence proved.

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