ICSE Class 9 Maths Selina Solutions Chapter 8 Logarithms

ICSE Class 9 Maths Selina Solutions Chapter 8: Here are the Selina answers to the problems found in the ICSE Class 9 Maths Selina Solutions Chapter 8 Logarithms. Students get in-depth knowledge on the subject of logarithms in this chapter. By completing all of the questions in the Selina textbook, students can easily receive a perfect score on their exams.

ICSE Class 9 Maths Selina Solutions Chapter 8 Overview

ICSE Class 9 Maths Selina Solutions Chapter 8 PDF

ICSE Class 9 Maths Selina Solutions Chapter 8 PDF

ICSE Class 9 Maths Selina Solutions Chapter 8 Exercise 8A

1. Express each of the following in logarithmic form:

(i) 5 ³ = 125

(ii) 3 ^-2 = 1/9

(iii) 10 ^-3 = 0.001

(iv) (81) ^3/4 = 27

Solution:

2. Express each of the following in exponential form:

(i) log ₈ 0.125 = -1

(ii) log ₁₀ 0.01 = -2

(iii) log _a A = x

(iv) log ₁₀ 1 = 0

Solution:

3. Solve for x: log ₁₀ x = -2.

Solution:

4. Find the logarithm of:

(i) 100 to the base 10

(ii) 0.1 to the base 10

(iii) 0.001 to the base 10

(iv) 32 to the base 4

(v) 0.125 to the base 2

(vi) 1/16 to the base 4

(vii) 27 to the base 9

(viii) 1/81 to the base 27

Solution;

5. State, true or false:

(i) If log ₁₀ x = a, then 10 ^x = a

(ii) If x ^y = z, then y = log _z x

(iii) log ₂ 8 = 3 and log ₈ 2 = 1/3

Solution:

6. Find x, if:

(i) log ₃ x = 0

(ii) log _x 2 = -1

(iii) log ₉ 243 = x

(iv) log ₅ (x – 7) = 1

(v) log ₄ 32 = x – 4

(vi) log ₇ (2x ² – 1) = 2

Solution:

7. Evaluate:

(i) log ₁₀ 0.01

(ii) log ₂ (1 ÷ 8)

(iii) log ₅ 1

(iv) log ₅ 125

(v) log ₁₆ 8

(vi) log _0.5 16

Solution:

8. If log _a m = n, express a ^{n – 1} in terms in terms of a and m.

Solution:

9. Given log ₂ x = m and log ₅ y = n

(i) Express 2 ^m-3 in terms of x

(ii) Express 5 ³ⁿ⁺² in terms of y

Solution:

10. If log ₂ x = a and log ₃ y = a, write 72 ^a in terms of x and y.

Solution:

11. Solve for x: log (x – 1) + log (x + 1) = log ₂ 1

Solution:

12. If log (x ² – 21) = 2, show that x = ± 11.

Solution:

ICSE Class 9 Maths Selina Solutions Chapter 8 Exercise 8B

1. Express in terms of log 2 and log 3:

(i) log 36

(ii) log 144

(iii) log 4.5

(iv) log 26/51 – log 91/119

(v) log 75/16 – 2log 5/9 + log 32/243

Solution:

2. Express each of the following in a form free from logarithm:

(i) 2 log x – log y = 1

(ii) 2 log x + 3 log y = log a

(iii) a log x – b log y = 2 log 3

Solution:

3. Evaluate each of the following without using tables:

(i) log 5 + log 8 – 2log 2

(ii) log ₁₀ 8 + log ₁₀ 25 + 2log ₁₀ 3 – log ₁₀ 18

(iii) log 4 + 1/3 log 125 – 1/5 log 32

Solution:

4. Prove that:

2log 15/18 – log 25/162 + log 4/9 = log 2

Solution:

5. Find x, if:

x – log 48 + 3 log 2 = 1/3 log 125 – log 3.

Solution:

6. Express log ₁₀ 2 + 1 in the form of log ₁₀ x.

Solution:

7. Solve for x:

(i) log ₁₀ (x – 10) = 1

(ii) log (x ² – 21) = 2

(iii) log (x – 2) + log (x + 2) = log 5

(iv) log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3

Solution:

8. Solve for x:

(i) log 81/log 27 = x

(ii) log 128/log 32 = x

(iii) log 64/log 8 = log x

(iv) log 225/log 15 = log x

Solution:

9. Given log x = m + n and log y = m – n, express the value of log 10x/y ² in terms of m and n.

Solution:

10. State, true or false:

(i) log 1 × log 1000 = 0

(ii) log x/log y = log x – log y

(iii) If log 25/log 5 = log x, then x = 2

(iv) log x × log y = log x + log y

Solution:

11. If log ₁₀ 2 = a and log ₁₀ 3 = b; express each of the following in terms of ‘a’ and ‘b’:

(i) log 12

(ii) log 2.25

(iii) log

(iv) log 5.4

(v) log 60

(iv) log

Solution:

12. If log 2 = 0.3010 and log 3 = 0.4771; find the value of:

(i) log 12

(ii) log 1.2

(iii) log 3.6

(iv) log 15

(v) log 25

(vi) 2/3 log 8

Solution:

13. Given 2 log ₁₀ x + 1 = log ₁₀ 250, find:

(i) x

(ii) log ₁₀ 2x

Solution:

14. Given 3log x + ½ log y = 2, express y in term of x.

Solution:

15. If x = (100) ^a , y = (10000) ^b and z = (10) ^c , find log 10√y/x ² z ³ in terms of a, b and c.

Solution:

16. If 3(log 5 – log 3) – (log 5 – 2 log 6) = 2 – log x, find x.

Solution:

ICSE Class 9 Maths Selina Solutions Chapter 8 Exercise 8(C)

1. If log ₁₀ 8 = 0.90; find the value of:

(i) log ₁₀ 4

(ii) log √32

(iii) log 0.125

Solution:

2. If log 27 = 1.431, find the value of:

(i) log 9 (ii) log 300

Solution:

3. If log ₁₀ a = b, find 10 ^{3b – 2} in terms of a.

Solution:

4. If log ₅ x = y, find 5 ^{2y+ 3} in terms of x.

Solution:

5. Given: log ₃ m = x and log ₃ n = y.

(i) Express 3 ^{2x – 3} in terms of m.

(ii) Write down 3 ^{1 – 2y + 3x} in terms of m and n.

(iii) If 2 log ₃ A = 5x – 3y; find A in terms of m and n.

Solution:

6. Simplify:

(i) log (a) ³ – log a

(ii) log (a) ³ ÷ log a

Solution:

7. If log (a + b) = log a + log b, find a in terms of b.

Solution:

8. Prove that:

(i) (log a) ² – (log b) ² = log (a/b). log (ab)

(ii) If a log b + b log a – 1 = 0, then b ^a . a ^b = 10

Solution:

9. (i) If log (a + 1) = log (4a – 3) – log 3; find a.

(ii) If 2 log y – log x – 3 = 0, express x in terms of y.

(iii) Prove that: log ₁₀ 125 = 3(1 – log ₁₀ 2).

Solution:

10. Given log x = 2m – n, log y = n – 2m and log z = 3m – 2n. Find in terms of m and n, the value of log x ² y ³ /z ⁴ .

Solution:

11. Given log _x 25 – log _x 5 = 2 – log _x 1/125; find x.

Solution:

ICSE Class 9 Maths Selina Solutions Chapter 8 Exercise 8D

1. If 3/2 log a + 2/3 log b – 1 = 0, find the value of a ⁹ .b ⁴

Solution:

2. If x = 1 + log 2 – log 5, y = 2log3 and z = log a – log 5; find the value of a if x + y = 2z.

Solution:

3. If x = log 0.6; y = log 1.25 and z = log 3 – 2log 2, find the values of:

(i) x + y – z        (ii) 5 ^{x + y – z}

Solution:

4. If a ² = log x, b ³ = log y and 3a ² – 2b ³ = 6 log z, express y in terms of x and z.

Solution:

5. If log (a – b)/2 = ½ (log a + log b), show that: a ² + b ² = 6ab.

Solution:

6. If a ² + b ² = 23ab, show that: log (a + b)/5 = ½ (log a + log b).

Solution:

7. If m = log 20 and n = log 25, find the value of x, so that: 2log (x – 4) = 2m – n.

Solution:

8. Solve for x and y; if x > 0 and y > 0; log xy = log x/y + 2log 2 = 2.

Solution:

9. Find x, if:

(i) log _x 625 = -4

(ii) log _x (5x – 6) = 2

Solution:

10. If p = log 20 and q = log 25, find the value of x, if 2log (x + 1) = 2p – q.

Solution:

11. If log ₂ (x + y) = log ₃ (x – y) = log 25/log 0.2, find the value of x and y.

Solution:

12. Given: log x/log y = 3/2 and log xy = 5; find the values of x and y.

Solution:

13. Given log ₁₀ x = 2a and log ₁₀ y = b/2

(i) Write 10 ^a in terms of x

(ii) Write 10 ^{2b + 1} in terms of y

(iii) If log ₁₀ p = 3a – 2b, express p in terms of x and y.

Solution:

14. Solve:

log ₅ (x + 1) – 1 = 1 + log ₅ (x – 1).

Solution:

15. Solve for x, if:

log _x 49 – log _x 7 + log _x 1/343 + 2 = 0

Solution:

16. If a ² = log x, b ³ = log y and a ² /2 – b ³ /3 = log c, find c in terms of x and y.

Solution:

17. Given: x = log ₁₀ 12, y = log ₄ 2 x log ₁₀ 9 and z = log ₁₀ 0.4, find

(i) x – y – z

(ii) 13 ^{x – y – z}

Solution:

18. Solve for x, log _x 15√5 = 2 – log _x 3√5

Solution:

19. Evaluate:

(i) log _b a x log _c b x log _a c

(ii) log ₃ 8 ÷ log ₉ 16

(iii) log ₅ 8/(log ₂₅ 16 x log ₁₀₀ 10)

Solution:

20. Show that:

log _a m ÷ log _ab m = 1 + log _a b

Solution:

21. If log _√27 x = 2 2/3, find x.

Solution:

22. Evaluate:

1/(log _a bc + 1) + 1/(log _b ca + 1) + 1/(log _c ab + 1)

Solution:

Benefits of ICSE Class 9 Maths Selina Solutions Chapter 8

1. Fundamental Understanding of Logarithms:

Concept Clarity: The ICSE Class 9 Maths Selina Solutions Chapter 8 introduces the fundamental concepts of logarithms, including the definition, properties, and rules. Understanding these basics is crucial as logarithms are foundational in many advanced mathematical concepts.

Building Blocks: ICSE Class 9 Maths Selina Solutions Chapter 8 helps students build a solid foundation for more complex topics in higher classes, including calculus and algebra.

2. Practical Applications:

Real-World Applications: Logarithms have real-world applications in fields like science (e.g., pH scale in chemistry), engineering, and computer science. Learning this chapter helps students understand how logarithms are used to solve practical problems.

Problem-Solving Skills: By solving a variety of problems, students improve their analytical and problem-solving skills, which are applicable in various scientific and technological fields.