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) 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 loga 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 {∵ loga 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: