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

**Exercise 8**

**Simplify the following (1 to 20):**

**1. (i) (81/16) ^{-3/4}**

**Solution**

**(i)** (81/16)^{-3/4}

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

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

= (3/2)^{-3/4 ×4}

= (3/2)^{-3}

= (2/3)^{3}

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

= (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 ^{-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**×10 ^{7}) (6×10^{-5})/(8×10^{10})**

**Solution**

**4. (i) 3a/b ^{-1} + 2b/a^{-1}**

**Solution**

3a/b^{-1} + 2b/a^{-1}

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

= (3a×b)/1 + (2b×a)/1

= 3ab + 2ab = 5ab

**(ii) 5 ^{0}×4^{-1} + 8^{1/3}**

**Solution**

5^{0}×4^{-1} + 8^{1/3}

= 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}

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

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

= (3/10)^{3× -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× -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

**Solution**

It can be written as

= (3)^{2n} ×(3)^{n}

= 3^{2n + n}

= 3^{3n}

= 100/600

= 1/6

**8. (i) [8 ^{-4/3 }÷ 2^{-2}]^{1/2}**

**(ii) (27/3)**

^{3 }- (1/4)^{-2 }÷ 5^{0}**Solution**

= (1/2)^{1}

= ½

^{2 }- (1/2)

^{-4}+ 1

^{4}+ 1

**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})**

**(ii) 1/(1+****a ^{m-n}) + **

**1/(1+**

**a**

^{n-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 ^{-1}y^{-1} + y^{-1}z^{-1} +z^{-1}x^{-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 = 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 **√(x/y).

**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