# 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 − 9x = 240

=> 9x (92 − 1) = 240

=> 9x = 240/80

=> 32x = 3

=> 2x = 1

=> x = 1/2

Therefore, (8x)x = [8 × (1/2)]1/2

= 41/2

= 2

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

Solution:

We have,

=> 3x+1 = 9x−2

=> 3x+1 = (32)x−2

=> 3x+1 = 32x−4

=> x + 1 = 2x − 4

=> x = 5

Therefore, 21+x = 21+5

= 26

= 64

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

Solution:

We are given,

=> 34x = (81)−1

=> 34x = (34)−1

=> 34x = (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 = 21+4(−1/4)

= 21−1

= 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 = 10−3

=> y = −3

Therefore, the value of x is 1 and the value of y is –3.

### (i) 3x+1 = 27 × 34

Solution:

We have,

=>  3x+1 = 27 × 34

=> 3x+1 = 33 × 34

=> 3x+1 = 37

=> x + 1 = 7

=> x = 6

### (ii) Solution:

We have,

=> => => => => 4x = −8/y = 3

=> x = 3/4 and y = −8/3

### (iii) 3x−1 × 52y−3 = 225

Solution:

We have,

=> 3x−1 × 52y−3 = 225

=> 3x−1 × 52y−3 = 32 × 52

=> x − 1 = 2 and 2y − 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 − 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) 4x−1 × (0.5)3−2x = (1/8)x

Solution:

We have,

=> 4x−1 × (0.5)3−2x = (1/8)x

=> (22)x−1 × (1/2)3−2x = [(1/2)3]x

=> 22x−2 × 22x−3 = 2−3x

=> 22x−2+2x−3 = 2−3x

=> 24x−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 = axb2y

=> a6/3 b−4/3 = axb2y

=> a2 b−4/3 = axb2y

=> x = 2 and 2y = −4/3

=> x = 2 and y = −2/3

### (i) , find x and y.

Solution:

We have,

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

=> (a−3 b6)7 ÷ (a5 b−8) = axby

=> (a−21 b42) ÷ (a5 b−8) = axby

=> (a−21−5 b42+8) = axby

=> (a−26 b50) = axby

=> x = −26, y = 50

### (ii) (a + b)−1(a−1 + b−1) = axby, find x+y+2.

Solution:

We have,

=> (a + b)−1(a−1 + b−1) = axby

=> = axby

=> = axby

=> 1/ab = axby

=> a−1b−1 = axby

=> x = −1 and y = −1

So, x+y+2 = −1−1+2 = 0.

### Question 19. If 2x × 3y × 5z = 2160, find x, y and z. Hence compute the value of 3x × 2−y × 5−z.

Solution:

We are given,

=> 2x × 3y × 5z = 2160

=> 2x × 3y × 5z = 24 × 33 × 51

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

Therefore, 3x × 2−y × 5−z = 34 × 2−3 × 5−1

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

= 81/40

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

Solution:

We are given,

=> 1176 = 2a × 3b × 7c

=> 23 × 31 × 72 = 2a × 3b × 7c

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

Therefore, 2a × 3b × 7−c = 23 × 31 × 7−2

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

= 24/49

### (i) Solution:

We have, = (xa+b−c)a−b (xb+c−a)b−c (xc+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 = xm+nyl, b = xn+lym and c = xl+myn, prove that am−n bn−l cl−m = 1.

Solution:

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

We have,

L.H.S. = am−n bn−l cl−m

= (xm+nyl)m−n(xn+lym)n−l(xl+myn)l−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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