ML Aggarwal Solutions for Class 9 Maths Chapter 8- Indices

Class 9: Maths Chapter 8 solutions. Complete Class 9 Maths Chapter 8 Notes.

ML Aggarwal Solutions for Class 9 Maths Chapter 8- Indices

ML Aggarwal 9th Maths Chapter 8, Class 9 Maths Chapter 8 solutions

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 + 8^1/3

Solution:

5⁰ 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

= (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 ÷ 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)

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

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 

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

Download PDF

ML Aggarwal Solutions for Class 9 Maths Chapter 8- Indices

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

Chapterwise ML Aggarwal Solutions for Class 9 Maths :

• Chapter 1- Rational and Irrational Numbers
• Chapter 2- Compound Interest
• Chapter 3- Expansions
• Chapter 4- Factorization
• Chapter 5- Simultaneous Linear Equations
• Chapter 6- Problems on Simultaneous Linear Equations
• Chapter 7- Quadratic Equations
• Chapter 8- Indices
• Chapter 9- Logarithms
• Chapter 10- Triangles
• Chapter 11- Mid Point Theorem
• Chapter 12- Pythagoras Theorem
• Chapter 13- Rectilinear Figures
• Chapter 14- Theorems on Area
• Chapter 15- Circle
• Chapter 16- Mensuration
• Chapter 17- Trigonometric Ratios
• Chapter 18- Trigonometric Ratios and Standard Angles
• Chapter 19- Coordinate Geometry
• Chapter 20- Statistics

About ML Aggarwal

M. L. Aggarwal, is an Indian mechanical engineer, educator. His achievements include research in solutions of industrial problems related to fatigue design. Recipient Best Paper award, Manipal Institute of Technology, 2004. Member of TSTE.