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

Here, we are providing the solutions for Chapter 8 Indices from ML Aggarwal Textbook for Class 9 ICSE Mathematics. Solutions of the second chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 9 Chapter 8 Indices ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on simplyfying equations and finding values.

### Exercise 8

Simplify the following (1 to 20):

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

Solution

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

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

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

= (3/2)-3/4 ×4

= (3/2)-3

= (2/3)3

= 23/33

= (2×2×2)/(3×3×3)

= 8/27

Solution

= (5/4)3× -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
×107) (6×10-5)/(8×1010)

Solution

4. (i) 3a/b-1 + 2b/a-1

Solution

3a/b-1 + 2b/a-1

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

= (3a×b)/1 + (2b×a)/1

= 3ab + 2ab = 5ab

(ii) 50×4-1 + 81/3

Solution

50×4-1 + 81/3

= 1 x (1/4) + (2)3× 1/3

= ¼ + 2

= (1 + 8)/4

= 9/4 = 2¼

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

Solution

(8/125)-1/3

= [(2×2×2)/(5×5×5)]-1/3

= (23/53)-1/3

= (2/5)3× -1/3

= (2/5)-1

= 5/2 = 2½

(ii) (0.027)-1/3

Solution

(0.027)-1/3

= (27/1000)-1/3

= [(3×3×3)/(10×10×10)]-1/3

= (33/103)-1/3

= (3/10)3× -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× -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

= 43×- 2/3 ÷ 32× -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

Solution

It can be written as

= (3)2n ×(3)n

= 32n + n

= 33n

= 100/600

= 1/6

8. (i) [8-4/3 ÷ 2-2]1/2

(ii) (27/3)- (1/4)-2 ÷ 50

Solution

= (1/2)1

= ½

= (3/2)- (1/2)-4 + 1
= 9/4 - (2)4 + 1
= 9/4 - 16 + 1
= 9/4 - 15
= -51/4

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)

(ii) 1/(1+am-n) + 1/(1+an-m)

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

(ii) (x+y+z)/(x-1y-1 + y-1z-1 +z-1x-1) = xyz

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 √(x/y).

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