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 8^{th} 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.

## ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices Download PDF

## 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}

= [(3^{4}/2^{4})]^{-3/4}

= [(3/2)^{4}]^{-3/4}

= (3/2)^{-3/4 x 4}

= (3/2)^{-3}

= (2/3)^{3}

= 2^{3}/3^{3}

= (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 ^{-3}b^{2})^{3}**

**Solution: **

(2a^{-3}b^{2})^{3}

= 2^{3} a ^{-3×3} b ^{2×3}

= 8a^{-1}b^{6}

**(ii) (a ^{-1} + b^{-1})/(ab)^{-1}**

**Solution: **

**3. (i) (x ^{-1} y^{-1})/(x^{-1} + y^{-1})**

**Solution:**

**(ii) (4 x 10 ^{7}) (6 x 10^{-5})/(8 x 10^{10})**

**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) 5 ^{0} x 4^{-1} + 8^{1/3}**

**Solution: **

5^{0} x 4^{-1} + 8^{1/3}

= 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}

= (2^{3}/5^{3})^{-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}

= (3^{3}/10^{3})^{-1/3}

= (3/10)^{3 x -1/3}

= (3/10)^{-1}

= 10/3

**6. (i) (-1/27) ^{-2/3}**

**Solution:**

(-1/27)^{-2/3}

= (-1/3^{3})^{-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

= (4^{3})^{-2/3} ÷ (3^{2})^{-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

= 1/4^{2} / 1/3^{3}

= 3^{3}/4^{2}

We get

= 27/16

= 1 11/16

**Solution:**

It can be written as

= (3)^{2n} × (3)^{n}

= 3 ^{2n + n}

= 3^{3n}

= 100/600

= 1/6

**Solution:**

= (1/2)^{1}

= ½

**9. (i) (3x ^{2})^{-3} × (x^{9})^{2/3}**

**(ii) (8x ^{4})^{1/3} ÷ x^{1/3}.**

**Solution:**

(i) (3x^{2})^{-3} × (x^{9})^{2/3}

We can write it as

(ii) (8x^{4})^{1/3} ÷ x^{1/3}

We can write it as

= 2 × x^{3/3}

So we get

= 2 × x^{1}

= 2 × x

= 2x

**10. (i) (3 ^{2})^{0} + 3^{-4} × 3^{6} + (1/3)^{-2}**

**(ii) 9 ^{5/2} – 3.(5)^{0} – (1/81)^{-1/2}**

**Solution:**

(i) (3^{2})^{0} + 3^{-4} × 3^{6} + (1/3)^{-2}

We can write it as

So we get

= 1 + 9 + 9

= 19

(ii) 9^{5/2} – 3.(5)^{0} – (1/81)^{-1/2}

We can write it as

Here

= 243 – 3 – (9 × 1)/1

= 240 – 9

= 231

**11. (i) 16 ^{3/4} + 2 (1/2)^{-1} (3)^{0}**

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

**Solution:**

(i) 16^{3/4} + 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} + 8^{0}]^{-1/2}**

**(ii) 3 ^{n} × 9^{n + 1} ÷ (3^{n – 1} × 9^{n – 1}).**

**Solution:**

(i) [(64)^{-2/3} 2^{-2} + 8^{0}]^{-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) 3^{n} × 9^{n + 1} ÷ (3^{n – 1} × 9^{n – 1})

We can write it as

= 3^{n} × (3 × 3)^{n + 1} ÷ (3^{n – 1} × (3 × 3)^{n – 1})

By further calculation

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

= 3^{n} × (3)^{2n + 2} ÷ (3^{n – 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

= 5^{6x – 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

= (x^{m – n})^{l}. (x^{n – 1})^{m}. (x^{1-m})^{n}

By further calculation

= (x)^{(m – n)l}. (x)^{(n – 1)m}. (x)^{(l – m)n}

= x^{ml – nl}. x^{nm – lm}. x^{ln – mn}

So we get

= x^{ml – nl + nm – lm + ln – mn}

= x^{0}

= 1

We can write it as

= (x^{a + b – c})^{a – b}. (x^{b + c – a})^{b – c}. (x^{c + 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

= x^{0}

= 1

**Solution:**

= x^{0}

= 1

= x^{0}

= 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 = c ^{z}, b = a^{x} and c = b^{y}, prove that xyz = 1.**

**Solution:**

It is given that

a = c^{z}, b = a^{x} and c = b^{y}

We can write it as

a = (b^{y})^{z} where c = b^{y}

So we get

a = b^{yz}

Here

a = (a^{x})^{yz}

a^{1} = a^{xyz}

By comparing both

xyz = 1

Therefore, it is proved.

**23. If a = xy ^{p – 1}, b = xy^{q – 1} and c = xy^{r – 1}, prove that**

**a ^{q – r}. b^{r – p}. c^{p – q} = 1.**

**Solution:**

It is given that

a = xy^{p – 1}

Here

a^{q – r} = (xy^{b – 1})^{q – r} = x^{q – r}. y^{(q – r) (p – 1)}

b = xy^{q – 1}

Here

b^{r – p} = (xy^{q – 1})^{r – p} = x^{r – p}. y^{(q – 1) (r – p)}

c = xy^{r – 1}

Here

c^{p – q} = (xy^{r – 1})^{p – q} = x^{p – q}. y^{(r – 1) (p – q)}

Consider

LHS = a^{q – r}. b^{r – p}. c^{p – q}

Substituting the values

= x^{q – r}. y^{(q – r) (p – 1)}. x^{r – p}. y^{(q – 1) (r – p)}. x^{p – q}. y^{(r – 1) (p – q)}

By further calculation

= x^{q – r + r – p – q}. y^{(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)}

So we get

= x^{0}. y^{pq – pr – q + r + qr – pr – r + p + rp – qr – p + q}

= x^{0}. y^{0}

= 1 × 1

= 1

= RHS

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

**Solution:**

Consider

2^{x} = 3^{y} = 6^{-z} = k

Here

2^{x} = k

We can write it as

2 = (k)^{1/x}

3^{y} = 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 2 ^{x} = 3^{y} = 12^{z}, prove that x = 2yz/y – z.**

**Solution:**

It is given that

2^{x} = 3^{y} = 12^{z}

Consider

2^{x} = 3^{y} = 12^{z} = k

Here

2^{x} = k where 2 = (k)^{1/x}

3^{y} = k where 3 = (k)^{1/y}

12^{z} = k where 12 = (k)^{-1/z}

We know that

12 = 2 × 2 × 3

Therefore, it is proved.

**26. Simplify and express with positive exponents:**

**(3x ^{2})^{0}, (xy)^{-2}, (-27a^{9})^{2/3}.**

**Solution:**

We know that

(3x^{2})^{0} = 1

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

**(i) a ^{a} + b^{b}**

**(ii) a ^{b} + b^{a}.**

**Solution:**

It is given that

a = 3 and b = – 2

(i) a^{a} + b^{b} = (3)^{3} + (-2)^{-2}

We can write it as

(ii) a^{b} + b^{a} = (3)^{-2} + (-2)^{3}

We can write it as

**28. If x = 10 ^{3} × 0.0099, y = 10^{-2} × 110, find the value of **

**Solution:**

It is given that

x = 10^{3} × 0.0099, y = 10^{-2} × 110

We know that

= √9

= √(3 × 3)

= 3

**29. Evaluate x ^{1/2}. y^{-1}. z^{2/3} when x = 9, y = 2 and z = 8.**

**Solution:**

It is given that

x = 9, y = 2 and z = 8

We know that

x^{1/2}. y^{-1}. z^{2/3} = (9)^{1/2}. (2)^{-1}. (8)^{2/3}

= 6

**30. If x ^{4}y^{2}z^{3} = 49392, find the values of x, y and z, where x, y and z are different positive primes.**

**Solution:**

It is given that

x^{4}y^{2}z^{3} = 49392

We can write it as

x^{4}y^{2}z^{3} = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 × 7

x^{4}y^{2}z^{3} = (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}) = p^{a}q^{b}, 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}) = p^{a}q^{b}

We can write it as

By cross multiplication

p^{-1}q^{-1} = p^{a}q^{b}

By comparing the powers

a = – 1 and b = – 1

Here

LHS = a + b + 2

Substituting the values

= – 1 – 1 + 2

= 0

= RHS