The availability of OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Ex 7(B) encourages students to tackle difficult exercises.

## 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) log_{3} 5 + log_{3} 2 + log_{3} 4

(iv) log_{2} 12 – log_{2} 2 + log_{2} 5

Solution:

Question 2.

Simplify without using tables :

(i) \(\frac{\log _{40} 1000}{\log _{40} 100}\)

(ii) \(\frac{\log 32}{\log 4}\)

(iii) log_{2 }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 m^{n} = 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 (2^{6} x 7)

= log 2^{6} + log 7

(∵ log mn = log n + log n and log m^{n} = 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) log_{10}n + log_{10}5 = 1

(ii) log_{3} n – log_{3} 4 = 2

(iii) log_{6} n – log_{6} (n – 1) = log_{6} 3

(iv) 2 log x = 4 log 3

Solution:

(i) log_{10} n + log_{10} 5 = 1

⇒ log_{10} (n x 5) = log_{10} 10

{∵ log m + log n = log mn and log_{a} a = 1}

⇒ log_{10} (5n) = log_{10} 10

Comparing, we get

5z = 10 ⇒ n = \(\frac { 10 }{ 5 }\) = 2

∴ n = 2

(ii) log_{3} n – log_{3} 4 = 2 = 2 x 1

⇒ log_{3} \(\frac { n }{ 4 }\) = 2 log_{3} 3 {∵ log_{a} a = 1}

⇒ log_{3} \(\frac { n }{ 4 }\) = log_{3} 3² = log_{3} 9

Comparing, we get

\(\frac { n }{ 4 }\) = 9 ⇒ n = 9 x 4 = 36

∴ n = 36

(iii) log_{6} n – log_{6} (n – 1) = log_{6} 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) log_{10} 5 + log_{10} 2

(ii) log_{10}4 + log_{10} 5 – log_{10} 2

(iii) 2 log_{10} 5 + log_{10} 8 – \(\frac { 1 }{ 2 }\) log_{10} 4

Solution: