# ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices is a useful resource for students to understand and prepare for the ICSE board examinations. These solutions help students obtain a clear conceptual knowledge of all the chapters of ML Aggarwal textbooks and also help them solve problems in an effective manner. Further, the experts at BYJU’S have provided the ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices PDF, in order to help students with their exam preparation.

The 8th chapter contains problems on simplifying the powers and exponents of numbers and algebraic expressions. The solutions can be utilised by students for any doubt clearance or quick reference during the process of self-study. The ML Aggarwal solutions are primarily created to help students with their exam preparations and boost their confidence in solving any difficult problems. All these solutions are according to the latest ICSE guidelines thus, ensuring the high possibilities of securing excellent marks in the examinations.

## Access ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices

Exercise 8

Simplify the following (1 to 20):

1. (i) (81/16)-3/4

Solution:

(81/16)-3/4

= [(34/24)]-3/4

= [(3/2)4]-3/4

= (3/2)-3/4 x 4

= (3/2)-3

= (2/3)3

= 23/33

= (2 x 2 x 2)/(3 x 3 x 3)

= 8/27

(ii)

Solution:

= (5/4)3 x -2/3

= (5/4)-2

= (4/5)2

= 16/25

2. (i) (2a-3b2)3

Solution:

(2a-3b2)3

= 23 a -3×3 b 2×3

= 8a-1b6

(ii) (a-1 + b-1)/(ab)-1

Solution:

3. (i) (x-1 y-1)/(x-1 + y-1)

Solution:

(ii) (4 x 107) (6 x 10-5)/(8 x 1010)

Solution:

4. (i) 3a/b-1 + 2b/a-1

Solution:

3a/b-1 + 2b/a-1

= 3a/(1/b) + 2b/(1/a)

= (3a x b)/1 + (2b x a)/1

= 3ab + 2ab = 5ab

(ii) 50 x 4-1 + 81/3

Solution:

50 x 4-1 + 81/3

= 1 x (1/4) + (2)3 x 1/3

= ¼ + 2

= (1 + 8)/4

= 9/4 = 2¼

5. (i) (8/125)-1/3

Solution:

(8/125)-1/3

= [(2 x 2 x 2)/(5 x 5 x 5)]-1/3

= (23/53)-1/3

= (2/5)3 x -1/3

= (2/5)-1

= 5/2 = 2½

(ii) (0.027)-1/3

Solution:

(0.027)-1/3

= (27/1000)-1/3

= [(3 x 3 x 3)/(10 x 10 x 10)]-1/3

= (33/103)-1/3

= (3/10)3 x -1/3

= (3/10)-1

= 10/3

6. (i) (-1/27)-2/3

Solution:

(-1/27)-2/3

= (-1/33)-2/3

= (-1/3)3 x -2/3

= (-1/3)-2

= (-3)2

= 9

(ii) (64)-2/3 ÷ 9-3/2

Solution:

(64)-2/3 ÷ 9-3/2

We can write it as

= (43)-2/3 ÷ (32)-3/2

By further calculation

= 4 3 ×- 2/3 ÷ 3 2 × -3/2

So we get

= 4-2 ÷ 3-3

= 4-2/ 3-3

It can be written as

= 1/42 / 1/33

= 33/42

We get

= 27/16

= 1 11/16

Solution:

It can be written as

= (3)2n × (3)n

= 3 2n + n

= 33n

= 100/600

= 1/6

Solution:

= (1/2)1

= ½

9. (i) (3x2)-3 × (x9)2/3

(ii) (8x4)1/3 ÷ x1/3.

Solution:

(i) (3x2)-3 × (x9)2/3

We can write it as

(ii) (8x4)1/3 ÷ x1/3

We can write it as

= 2 × x3/3

So we get

= 2 × x1

= 2 × x

= 2x

10. (i) (32)0 + 3-4 × 36 + (1/3)-2

(ii) 95/2 – 3.(5)0 – (1/81)-1/2

Solution:

(i) (32)0 + 3-4 × 36 + (1/3)-2

We can write it as

So we get

= 1 + 9 + 9

= 19

(ii) 95/2 – 3.(5)0 – (1/81)-1/2

We can write it as

Here

= 243 – 3 – (9 × 1)/1

= 240 – 9

= 231

11. (i) 163/4 + 2 (1/2)-1 (3)0

(ii) (81)3/4 – (1/32)-2/5 + (8)1/3 (1/2)-1 (2)0.

Solution:

(i) 163/4 + 2 (1/2)-1 (3)0

We can write it as

So we get

= (2)3 + 4

= 2 × 2 × 2 + 4

= 8 + 4

= 12

(ii) (81)3/4 – (1/32)-2/5 + (8)1/3 (1/2)-1 (2)0

We can write it as

= 27 – 4 + 4

= 27

Solution:

= 9/4

= 2 ¼

= 19

13. (i) [(64)-2/3 2-2 + 80]-1/2

(ii) 3n × 9n + 1 ÷ (3n – 1 × 9n – 1).

Solution:

(i) [(64)-2/3 2-2 + 80]-1/2

We can write it as

= [4 × 1 × 1]-1/2

= (4)-1/2

Here

= (2 × 2)-1/2

= (2)2 × -1/2

= (2)-1

= 1/(2)1

= ½

(ii) 3n × 9n + 1 ÷ (3n – 1 × 9n – 1)

We can write it as

= 3n × (3 × 3)n + 1 ÷ (3n – 1 × (3 × 3)n – 1)

By further calculation

= 3n × (3)2 × (n + 1) ÷ (3n – 1 × (3)2(n-1)])

= 3n × (3)2n + 2 ÷ (3n – 1 × (3)2n – 2)

So we get

= (3)n + 2n + 2 ÷ (3)n – 1 + 2n – 2

= (3)3n + 2 ÷ (3)3n – 3

Here

= (3)3n + 2 – 3n + 3

= (3)5

We get

= 3 × 3 × 3 × 3 × 3

= 243

Solution:

= 2 – 4

= – 2

Solution:

= 4

= 56x – 2 – 6x

= 5-2

= 1/(5)2

= 1/25

Solution:

= 7 – 7 × 7

= 7 – 49

= – 42

(ii) (27)4/3 + (32)0.8 + (0.8)-1

We can write it as

= 98.25

Solution:

= (3)1

= 3

Solution:

We can write it as

= (xm – n)l. (xn – 1)m. (x1-m)n

By further calculation

= (x)(m – n)l. (x)(n – 1)m. (x)(l – m)n

= xml – nl. xnm – lm. xln – mn

So we get

= xml – nl + nm – lm + ln – mn

= x0

= 1

We can write it as

= (xa + b – c)a – b. (xb + c – a)b – c. (xc + a – b)c – a

By further calculation

= x(a + b – c) (a – b). x(b + c – a) (b – c). x(c + a – b) (c – a)

So we get

= x0

= 1

Solution:

= x0

= 1

= x0

= 1

20. (i) (a-1 + b-1) ÷ (a-2 – b-2)

Solution:

(i) (a-1 + b-1) ÷ (a-2 – b-2)

We can write it as

= 1

21. Prove the following:

(i) (a + b)-1 (a-1 + b-1) = 1/ab

Solution:

(i) (a + b)-1 (a-1 + b-1) = 1/ab

Here

LHS = (a + b)-1 (a-1 + b-1)

We can write it as

= RHS

Hence, proved.

= xyz

= RHS

Hence, proved.

22. If a = cz, b = ax and c = by, prove that xyz = 1.

Solution:

It is given that

a = cz, b = ax and c = by

We can write it as

a = (by)z where c = by

So we get

a = byz

Here

a = (ax)yz

a1 = axyz

By comparing both

xyz = 1

Therefore, it is proved.

23. If a = xyp – 1, b = xyq – 1 and c = xyr – 1, prove that

aq – r. br – p. cp – q = 1.

Solution:

It is given that

a = xyp – 1

Here

aq – r = (xyb – 1)q – r = xq – r. y(q – r) (p – 1)

b = xyq – 1

Here

br – p = (xyq – 1)r – p = xr – p. y(q – 1) (r – p)

c = xyr – 1

Here

cp – q = (xyr – 1)p – q = xp – q. y(r – 1) (p – q)

Consider

LHS = aq – r. br – p. cp – q

Substituting the values

= xq – r. y(q – r) (p – 1). xr – p. y(q – 1) (r – p). xp – q. y(r – 1) (p – q)

By further calculation

= xq – r + r – p – q. y(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)

So we get

= x0. ypq – pr – q + r + qr – pr – r + p + rp – qr – p + q

= x0. y0

= 1 × 1

= 1

= RHS

24. If 2x = 3y = 6-z, prove that 1/x + 1/y + 1/z = 0.

Solution:

Consider

2x = 3y = 6-z = k

Here

2x = k

We can write it as

2 = (k)1/x

3y = k

We can write it as

3 = (k)1/y

6-z = k

We can write it as

6 = (k)-1/z

So we get

2 × 3 = 6

(k)1/x × (k)1/y = (k)-1/z

By further calculation

(k)1/x + 1/y = (k)-1/z

We get

1/x + 1/y = – 1/z

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

Therefore, it is proved.

25. If 2x = 3y = 12z, prove that x = 2yz/y – z.

Solution:

It is given that

2x = 3y = 12z

Consider

2x = 3y = 12z = k

Here

2x = k where 2 = (k)1/x

3y = k where 3 = (k)1/y

12z = k where 12 = (k)-1/z

We know that

12 = 2 × 2 × 3

Therefore, it is proved.

26. Simplify and express with positive exponents:

(3x2)0, (xy)-2, (-27a9)2/3.

Solution:

We know that

(3x2)0 = 1

27. If a = 3 and b = – 2, find the values of:

(i) aa + bb

(ii) ab + ba.

Solution:

It is given that

a = 3 and b = – 2

(i) aa + bb = (3)3 + (-2)-2

We can write it as

(ii) ab + ba = (3)-2 + (-2)3

We can write it as

28. If x = 103 × 0.0099, y = 10-2 × 110, find the value of

Solution:

It is given that

x = 103 × 0.0099, y = 10-2 × 110

We know that

= √9

= √(3 × 3)

= 3

29. Evaluate x1/2. y-1. z2/3 when x = 9, y = 2 and z = 8.

Solution:

It is given that

x = 9, y = 2 and z = 8

We know that

x1/2. y-1. z2/3 = (9)1/2. (2)-1. (8)2/3

= 6

30. If x4y2z3 = 49392, find the values of x, y and z, where x, y and z are different positive primes.

Solution:

It is given that

x4y2z3 = 49392

We can write it as

x4y2z3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 × 7

x4y2z3 = (2)4 (3)2 (7)3 ……. (1)

Now compare the powers of 4, 2 and 3 on both sides of equation (1)

x = 2, y = 3 and z = 7

31. If , find x and y, where a, b are different positive primes.

Solution:

It is given that

By comparing the base on both sides

2 = x

x = 2

– 4/3 = 2y

2y = – 4/3

By further calculation

y = – 4/3 × ½ = – 2/3

32. If (p + q)-1 (p-1 + q-1) = paqb, prove that a + b + 2 = 0, where p and q are different positive primes.

Solution:

It is given that

(p + q)-1 (p-1 + q-1) = paqb

We can write it as

By cross multiplication

p-1q-1 = paqb

By comparing the powers

a = – 1 and b = – 1

Here

LHS = a + b + 2

Substituting the values

= – 1 – 1 + 2

= 0

= RHS