Chapter 8 Indices ML Aggarwal ICSE Solutions for Class 9 Maths

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

(ii)

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 + 8^1/3

Solution

5⁰×4^-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

= (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

= 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

Solution

It can be written as

= (3)²ⁿ ×(3)ⁿ

= 3^{2n + n}

= 3³ⁿ

= 100/600

= 1/6

8. (i) [8^-4/3 ÷ 2^-2]^1/2

(ii) (27/3)³ - (1/4)^-2 ÷ 5⁰

Solution

= (1/2)¹

= ½

= (3/2)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 ÷ x^1/3

Solution

(i) (3x²)^-3 × (x⁹)^2/3

We can write it as

(ii) (8x⁴)^1/3 ÷ x^1/3

We can write it as

= 2 × x^3/3

So we get

= 2 × x¹

= 2 × x

= 2x

10. (i) (3²)⁰ + 3^-4×3⁶ + (1/3)^-2

(ii) 9^5/2 – 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) 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 + 8⁰]^-1/2

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

We can write it as

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

By further calculation

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

= 3ⁿ × (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}

= x⁰

= 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,

= x⁰

= 1

Solution

= x⁰

= 1

= x⁰

= 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^-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 = 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

⇒ a¹ = 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

= x⁰. y^{pq – pr – q + r + qr – pr – r + p + rp – qr – p + q}

= x⁰. y⁰

= 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:

(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) 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 = 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 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 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) = p^aq^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^aq^b

We can write it as

By cross multiplication

p^-1q^-1 = p^aq^b

By comparing the powers

a = –1 and b = –1

Here,

LHS = a + b + 2

Substituting the values

= –1 –1 + 2

= 0

= RHS