OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Ex 7(B)

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S Chand Class 9 ICSE Maths Solutions Chapter 7 Logarithms Ex 7(B)

Question 1.
Express each sum or difference as a single logarithm :
(i) log 6 + log 5
(ii) log 12 – log 2
(iii) log₃ 5 + log₃ 2 + log₃ 4
(iv) log₂ 12 – log₂ 2 + log₂ 5
Solution:

Question 2.
Simplify without using tables :
(i) \(\frac{\log _{40} 1000}{\log _{40} 100}\)
(ii) \(\frac{\log 32}{\log 4}\)
(iii) log₂ 8
Solution:

Question 3.
Simplify:
(i) log (m²) – log m
(ii) log y² ÷ log y
(iii) log 24 – log 3
(iv) log 32 + log 4 – log 16
(v) log 256 – log 1024
(vi) log 256 – log 1024
Solution:

Question 4.
Prove that:
(i) log 7 + log \(\frac { 1 }{ 7 }\) = 0
(ii) log 72 = 3 log 2 + 2 log 3
(iii) log 448 = 6 log 2 + log 7
(iv) log \(\frac { 4 }{ 7 }\) + log \(\frac { 33 }{ 18 }\) – log \(\frac { 22 }{ 21 }\) = 0
(v) log \(\sqrt[3]{6 \frac{2}{9}}\) = \(\frac { 1 }{ 3 }\)(log 2 – 2 log 3) + log 2
(vi) (log a)² – (log b)² = log (ab) log \(\frac { a }{ b }\)
Solution:
(i) log 7 + log \(\frac { 1 }{ 7 }\) = 0
L.H.S. = log 7 + log \(\frac { 1 }{ 7 }\) = log 7 x \(\frac { 1 }{ 7 }\) (∵ log mn = log m + log n)
= log 1 = 0 = R.H.S. (∵ log 1 = 0)

(ii) log 72 = 3 log 2 + 2 log 3
L.H.S. = log 72 = log (2³ x 3²)
= log 2³ + log 3²
(∵ log mn = log n + log n and log mⁿ = n log m)

72 = 2³ x 3²
= 3 log 2 + 2 log 3 = R.H.S

(iii) log 448 = 6 log 2 + log 7

L.H.S. = log 448 = log (2⁶ x 7)
= log 2⁶ + log 7
(∵ log mn = log n + log n and log mⁿ = n log m)
6 log 2 + log 7 = R.H.S

(iv) \(\log \frac{4}{7}+\log \frac{33}{18}-\log \frac{22}{21}\) = 0

Question 5.
Solve:
(i) log₁₀n + log₁₀5 = 1
(ii) log₃ n – log₃ 4 = 2
(iii) log₆ n – log₆ (n – 1) = log₆ 3
(iv) 2 log x = 4 log 3
Solution:
(i) log₁₀ n + log₁₀ 5 = 1
⇒ log₁₀ (n x 5) = log₁₀ 10
{∵ log m + log n = log mn and log_a a = 1}
⇒ log₁₀ (5n) = log₁₀ 10
Comparing, we get
5z = 10 ⇒ n = \(\frac { 10 }{ 5 }\) = 2
∴ n = 2

(ii) log₃ n – log₃ 4 = 2 = 2 x 1
⇒ log₃ \(\frac { n }{ 4 }\) = 2 log₃ 3 {∵ log_a a = 1}
⇒ log₃ \(\frac { n }{ 4 }\) = log₃ 3² = log₃ 9
Comparing, we get
\(\frac { n }{ 4 }\) = 9 ⇒ n = 9 x 4 = 36
∴ n = 36

(iii) log₆ n – log₆ (n – 1) = log₆ 3

(iv) 2 log x = 4 log 3
⇒ log x = 2 log 3 (Dividing by 2)
⇒ log x = log 3² = log 9
Comparing, we get
x = 9

Question 6.
Simplify : (Do not use tables)
(i) log₁₀ 5 + log₁₀ 2
(ii) log₁₀4 + log₁₀ 5 – log₁₀ 2
(iii) 2 log₁₀ 5 + log₁₀ 8 – \(\frac { 1 }{ 2 }\) log₁₀ 4
Solution: