# OP Malhotra Class 9 ICSE Maths Solutions Chapter 7

## OP Malhotra Class 9 ICSE Maths Solutions Chapter 7 Logarithms

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 7, Logarithms. Here students can easily find step by step solutions of all the problems for Logarithms, Exercise 7a, 7b and 7c Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 7 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Logarithms Exercise 7(a) Solution :

Question no – (1)

Solution :

(i) log2 8 = 3

log2 8 = 3

= 8 = 23

Thus, Exponential form log2 8 = 3 is 8 = 23

(ii) log3 81 = 4

log3 81 = 4

= 81 = 34

Hence, Exponential form of log3 81 = 4 is 81 = 34

(iii) log2 1/2 = -1

log2 1/2 = -1

= 2-1 = 1/2

Thus, Exponential form log2 1/2 = -1 is 2-1 = 1/2

(iv) log5 1/25 = -2

log5 1/25 = -2

= 1/25 = 5-2

Hence, Exponential form of log5 1/25 = -2 is 1/25 = 5-2

(v) 1/2 = log4 2

1/2 = log4 2

= 2 = 41/2

Thus, Exponential form 1/2 = log4 2 is 2 = 41/2

(vi) 1/3 = log27 3

1/3 = log27 3

= 271/2 = 3

Therefore, exponential form 1/3 = log27 3 is 271/2 = 3

Question no – (2)

Solution :

(i) Given, 16 = 24

16 = 24

= log2 16 = 4

∴ Logarithmic form of 16 = 2is log2 16 = 4

(ii) Given, 25 = 52

= 25 = 52

= log5 25 = 2

∴ Logarithmic form of 25 = 52 is log5 25 = 2

(iii) Given, 81 = 34

= 81 = 34

= log3 81 = 4

∴ Logarithmic form of 81 = 34 is log3 81 = 4

(iv) Given, 60 = 1

= 60 = 1

= log6 = 0

∴ Logarithmic form of 60 = 1 is log6 = 0

(v) Given, 81/3 = 2

= 81/3 = 2

= log8 2 = 1/3

∴ Logarithmic form of 81/3 = 2 is log8 2 = 1/3

(vi) Given, 1/9 = 3-2

= 1/9 = 3-2

= log3 1/9 = -2

Logarithmic form of 1/9 = 3-2 is log3 1/9 = -2

(vii) 1/32 = 2-5

= 1/32 = 2-5

= log2 1/32 = -5

∴ Logarithmic form of 1/32 = 2-5 is log2 1/32 = -5

(viii) 1/32 = 2-5

In Logarithmic form = log2 1/32 = -5

Question no – (3)

Solution :

(i) Given, log10 1000

= Let, log10 1000 = x

= 10x = 1000

= 103

x = 3

So, the value will be 3

(ii) log2 8

= Let, log2 8 = x

= 2x = 8

= 23

x = 3

Thus, the value will be 3

(iii) log3 81

= Let, log3 81 = x

= 3x = 81

= 3x = 34

= x = 4

Thus, the value will be 4

(iv) log10 0.1

= Let, log10 0.1 = x

= 10x = 0.1

= 1/10

= 10x = 10-1

= x = -1

Thus, the value will be -1

(v) In the question we get, log10 0.01

∴ log10 0.01

= 10x = 1/100

= 1/102

= 10x = 10-2

= x = -2

Hence, the value of log10 0.01 is -2

(vi) log10 0.0001

Now, log10 0.0001

= 102 = 0.0001 = 1/10000

= 10x = 1/104

= 10x = 10-4

= x = -4

Thus, the value of log10 0.0001 is -4

(vii) log2 1/4

∴ log2 1/4

= 2x = 1/4

= 2x = 2-2

= x = -2

Thus, the value will be -2

(viii) log3 1/27

= 3x = 1/27

= 1/33

= 3x = 3-3

1x = -31

Hence, the value is -31

(ix) log3 1

= 3x = 1

= 3x = 30

= x = 0

So, the value is 0

(x) log1/2 1/4

= (1/2)x = 1/4

= (1/2)2

= x = 2

Therefore, the value will be 2

(xi) log27 9

= 27x = 9

= (3)3x = 32

= 3x = 2

= x = 2/3

Thus, the value will be 2/3

(xii) log1/5 125

= (1/5)x = 125

= 53

= (1/5)x = (1/5)-3

= x = -3

Hence, the value is -3

(xiii) log5/6 1

= (5/6)x = 1

= (5/6)0

= x = 0

Thus, the value is 0

(xiv) log1/3 9

= (1/3)x = 9

= (3)2

= (1/3)-2

= x = -2

So, the value is -2

(xv) log10 10

= 10x = 10

= 101

= x = 1

Thus, the value will be 1

Question no – (4)

Solution :

(i) logx 216 = 3

= x3 = 216

= x3 = 63

= x = 6

Thus, the value of x will be 6

(ii) log4 x = -4

= 4 -4 = x

= x = 4 -4

= x = 1/44

= x = 1/256

Thus, the value of x will be 1/256

(iii) , log3 x = 0

∴ log3 x = 0

= 3 = x

= 1 = x

Therefore, the value of x will be 1

(iv) log8 x = 2/3

= 8 2/3 = x

= (232/3 = x

= 22 = x

= 4 = x

Hence, the value of x is 4

(v) log10 100 = x

= 10x = 100

= 102

= x = 2

Thus, the value of x will be 2

(vi) log10 100 = x

= 10x = 100

= 102

= x = 2

Thus, the value of x will be 2

Question no – (5)

Solution :

Given, If log10 x = a, then 10a = x.

= Then,

= 10x = x

Therefore, we can say, It is True.

Logarithms Exercise 7(b) Solution :

Question no – (1)

Solution :

(i) log6 + log 5

= log (6 × 5)

= log 30

(ii) log 12 + log 2

∴ log 12 + log 2

= log 12/2

= log 6

(iii) log3 5 + log3 2 + log3 4

Now, log3 5 + log3 2 + log3 4

= log3 (5 × 2 ×4)

= log3 40

(iv) log2 12 – log2 2 + log2 2 + log2 5

Now, log2 12 – log2 2 + log2 2 + log2 5

= log2 (12/2 × 5)

= log2 30

Question no – (2)

Solution :

(i) log40 1000/log40 100

= log40 103/log40 102

= 3log40 10/2 log40 10

= 3/2

(ii) log32/log4

= log25/log22

= 5 log 2/2log 2

= 5/2

(iii) log2 8

= log28 = log2 23

= 3log22

= 3 × 1

= 3

Question no – (3)

Solution :

(i) log (m2) – log m

∴ log (m2) – log m

= log(m2/m)

= log m

(ii) logy2 ÷ log y

∴ logy2 ÷ log y

= logy2/logy

= 2 logy/logy

= 2

(iii) log 24 – log 3

Now, log 24 – log 3

= log 24/3

= log 8

= log (2)3

= 3 log 2

(iv) In the question we get, log 32 + log 4 – log 16

log 32 + log 4 – log 16

= log (32 × 410)

= log 8

= log (2)3

= 3 log 2

(v) log 256 – log 1024

log 256 – log 1024

= log (256/1024)

= log (1/4)

= log (1/22)

= log (2)-2

= – 2log 2

(vi) log 256 ÷ log 1024

log 256 ÷ log 1024

= log 256/log 1024

= log28/log210

= 8 log 2/10 log 2

= 8/10

= 4/5

Question no – (4)

Solution :

As per the given question,

(i) log 7 + log 1/7 = 0

L.H.S

log (7 × 1/7)

= log 1

= 0

= L.H.S = R.H.S …(Proved)

(ii) log 72 = 3 2 + 2 log 3

L.H.S

= log (8 × 9)

= log (23 × 32)

= 3 log 2 + 2 log 3

∴ L.H.S = R.H.S …(Proved)

(iii) log 448= 6 log 2 + log 7 …(according to the question)

L.H.S

= log (64 × 7)

= log 64 + log 7

= log 26 + log 7

= log 26 + log 7

= 6 log 2 + log 7

L.H.S  = R.H.S …(Proved)

(iv) log 4/7 + log 33/18 – log 22/21 = 0

L.H.S

= log 4/7 + log 33/18 – log 22/21

= log (4/7 × 33/18/22/21)

= log (4/7 × 33/18 × 21/22)

= log (1)

= 0

Therefore, L.H.S = R.H.S …(Proved)

(v) log 3√6 2/9 = 1/3 (log 7 – 2 log 3) + log 2 ……(according to question)

L.H.S log 3√56/9

= log (56/9)1/3

= 1/3 [log 56 – log9]

= 1/3 [log (8 × 7) – log 32]

= 1/3 log 8 + 1/3 log 7 – 1/3 × 2 log 3 .

∴ L.H.S = R.H.S ….(Proved)

(vi) (log a)2 – (log b)2 = log (ab) log (a/b) …(according to the question)

L.H.S, (log a)2 – (log b)2

= log (ab). log (a/b)

Thus, L.H.S = RHS …(Proved)

Question no – (5)

Solution :

(i) log10 n + log10 5 = 1

∴ log10 n + log10 5 = 1

= log 10 (n × 5) = log10 10

= 5n = 10

= x = 10/5

= x = 2 …(Solved)

(ii) log3 n – log3 4 = 2

log3 n – log3 4 = 2

= log3n – log3y = log39

= log3 (n/4) = log39

= n/4 = 9

= n = 36 …(Solved)

(iii) log6 n – log6 (n – 1) = log6 3

Now, log6 n – log6 (n – 1) = log6 3

= log6 n/n-1 = log63

= n/n-1 = 3

= n = 3n – 3

= – 2x = – 3

= n = 3/2 …(Solved)

Question no – (6)

Solution :

According to the question,

(i) log10 5 + log10 2

∴ log10 5 + log10 2

= log10 (5 × 2)

= log10 (10)

= 1 …(Simplified)

(ii) log10 4 + log10 5 – log10 2

∴ log10 4 + log10 5 – log10 2

= log10 (4 × 5/2)

= log10 10

= 1 …(Simplified)

(iii) 2 log10 5 + log10 8 – 1/2 log10 4

∴ 2 log10 5 + log10 8 – 1/2 log10 4

= log10 52 + log10 8 – log10 4 1/2

= log1025 + log10 8 – log10 2

= log 25 × 8/2

= log10 100

= log10 102

= 2 log10 10

= 2 × 1

= 2 …(Simplified)

Logarithms Exercise 7(c) Solution :

From the given question we know,

log 2 = 0.3010,

log 3 = 0.4771,

log 5 = 0.6990,

log 7 = 0.8451

log 11 = 1.0414

Question no – (1)

Solution :

Now, the value of log 6,

= log (3 × 2)

= log3 + log2

= 0.4771 + 0.3010

= 0.7781

Thus, the value of log 6 will be 0.7781

Question no – (2)

Solution :

Now, the value of log 12,

= log (4 × 3)

= log 4 + log 3

= log 22 + log 3

= 2 log 2 + log 3

= 2 × 0.3010 + 0.4771
= 0.6020 + 0.4771

= 1.00791

Thus, the value of log 12 will be 1.00791

Question no – (3)

Solution :

Now, the value of log 15

= log (3 × 5)

= log 3 + log 5

= 0.4771 + 0.6990

= 1.1761

Hence, the value of log 15 = 1.1761

Question no – (5)

Solution :

Now, the value of log 36,

= log (4 × 9)

= log 4 + log 9

= log 22 + log32

= 2 log 2 + 2 log 3

= 2 × 0.3010 + 2 × 0.4771

= 1.5562

Therefore, the value of log 36 will be 1.5562

Question no – (6)

Solution :

Now, the value of log 80,

= log (8 × 10)

= log 8 + log10

= log23 + log10

= 3 log 2 + ,log 10

= 3 × 0.3010 + 1.0000

= 1.9030

Thus, the value of log 80 will be 1.9030

Question no – (7)

Solution :

Now, the value of log 2 1/3

= log 7/3

= log 7 – log 3

= 0.8451 – 0.4771

= 0.3680

Thus, the value of log 2 1/3 = 0.3680

Question no – (8)

Solution :

Now, the value of log 113

= 3 log 11

= 3 × 1.0414

= 3.1242

Therefore, the value of log 11= 3.1242

Question no – (9)

Solution :

Now, the value of log (2 1/3)5

= 5 log (2 1/3)

= 8 log (7/3)

= 5 [log 7 – log3]

= 5 [0.8451 – 0.471]

= 5 × 0.3680

= 1.8400

Therefore, the value of log (2 1/3)5 will be 1.8400

Question no – (10)

Solution :

As per the question, log 6 = 0.7782,

Now, the value of log 36.

= log 3.6 = log (6)2

= 2 log 6

= 2 × 0.7782

= 1.5564

Hence, the value of log 36 will be 1.5564

Question no – (11)

Solution :

(i) log103,

= log10 (45/25)

= log10 45 – log10 25

= y – x

(ii) log102

= log10 (10/5)

= log10 10 – log105

= 1 – x/2

= 2 – x/2

Question no – (12)

Solution :

As per the question,

(i) log 31.87 = x

Now, log (31.87)2, = x

= log (31.87)2,

= 2 log (31.87)

= 2x

(ii) log 31.87 = x

Now, log10 0.03187,

= log10 (0.03187)

= log (31.87/1000)

= log 31.87 – log1000

= x – 3

(iii) log 31.87 = x

Now, log10 √31870

= log10 (31870)1/2

= 1/2 log 10 (31.87 × 100)

= 1/5 (log 31.87 + log10 1000)

= 1/5 [log 31.87 + 3 log10 10]

= 1/2 (x + 3)

Question no – (13)

Solution :

(i) According to the question,

log10 (x + 1) + log10 (x – 1) = log10 11 – 1 log10 32

log10 (x + 1) + log10 (x – 1) = log10 11 – 1 log10 32

= log10 + (x + 1) (x – 1) = log10 (11 × 9)

= x2 – 1 = 99

= x2 = 99 + 1

= 100

= x = ± 10 (Solved)

(ii) In the given question, log (10x + 5) – log (x – 4) = 2

Now, log (10x + 5) – log (x – 4) = 2

= log (10x + 5/x – 4) = 2 log10 10

= 10x + 5/x – 4 = 102

= 10x + 5 = 100 (x – 4)

= 10x + 5 100x – 400

= 90x = 405

= x = 405/90

= 4.5

Question no – (14)

Solution :

(i) 2 log10 x + 1/2 log10 y = 1 …(as per the question)

= log10x2 + log10 y 1/2 = log10 10

= log (x2√y) = log10 (10)

= x2 √y = 10
= √y = 10/x2

y = 100/x4

(ii) 2 log 3 – 1/2 log 16 + log 12 ……(according to the question)

Now, 2 log 3 – 1/2 log 16 + log 12.

= log32 – log (16)1/2 + log 12

= log 9 – log 4 + log 12

= log (9 × 12/4)

= log 27

Question no – (15)

Solution :

Given, log10 y + 2 log10x = 2

Now, log10 y + 2 log10x = 2

= log10y + log10 x2 = 2 log10 10

= y × x2 = 100

= y = 100/x2

y = 100 x-2

Question no – (16)

Solution :

According to the question we know,

a = 1 + log10 2 – log10 5,

b = 2 log10 3,

c = log10 m – log10 5

Now, the value of m,

a = log10 10 + log10 2 – log105

= log10 (10 × 2/5)

= log10 4

and, b = 2 log10 3

= log 32

= log 109

C = log10 m – log10 5

= log10 (m/5)

a + b + c = 2c

= log104 + log10 9 = 2 log (m/5)

= log10 (9 × 4) = log10 (m/5)2

= 36 = m2/25
= m2 = 25 × 35

= m = 5 × 6

= m = 5 × 6

= m  = 3

Therefore, the value of m will be 3

Question no – (17)

Solution :

Given in the question, 2 + 1/2 log10 9 – 2 log10 5

Now, 2 log10 10 + 1/2 log10 9 – 2 log10 5

= log102 + log10 3 – log10 (5)2

= log102 + log10 3 – log10 25

= log10 (100 × 3/25)

= log10 12

Question no – (18)

Solution :

As per the question we know,

a = log 12,

b = log 6 and

c = 2 log √2

(i) Now, a – b – c

= log 12 – log 6 – 2 log  √2

= log (12/6) – log (2)1/2 × 2

= log 2 – log 2

= 0

Therefore, a – b – c = 0

(ii) Now, 9a-b-c

= 9°

= 1
Therefore, 9a-b-c = 0

Question no – (19)

Solution :

As per the question the value of,

x = log10 12,

y = log2 × log10 9

z = log10 (0.4)

(i) Now, x – y – z,

= log10 12 – log42 × log42 × log10 9 – log10 (0.4)

= log10 54

= log10 (12/0.4) – log4 (4)1/2 × log 310 2

= log10 (30) – 1/2 = log10 9

= log10 (30) – log103.

= log10 30/3

= log10 10

= 1

Question no – (20)

Solution :

In the given question the value of,

p = log10 20

q = log10 25

log10 (x + 1)2 = 2 log10 20 – log10 25

= log (x + 1) = log10 202 – log1025

= log10 400 – log10 25

= log10 (400/25)

= log10 16

= log 10 42

= (x + 1)2 = 42

= (x + 1) = 4

= x = 4 – 1

= 3

Question no – (21)

Solution :

Given, 3 + log10 (10-2)

= 3 log10 10 + (-2) log10 10

= 3 log 10 10 – 2 log10 10

= log10 10

= 1

Question no – (22)

Solution :

(i) log10 x = a, log 10y = b,

= 10a = 10a/10 = x/10

= [log 10x = a

(ii) log10 x = a, log 10y = b,

102b = (10b)2 = y2

= [log10 y = b]

= 10b = y

(iii) log10 x = a, log 10y = b,

Now, log10 p = 2a – b

= 2 log10 x – log10y

= log10x2 – log10y

= p x2/y

Question no – (23)

Solution :

In the question we get,

2 log10 5 + log10 + 8 – 1/2 log10 4.

2 log10 5 + log10 + 8 – 1/2 log10 4.

= log1052 + log10 8 – log10 (4)1/2

= log10 25 + log108 – log102

= log1025 + log108 – log 102

= log10 (25 × 8/2)

= log10 100

= log10 102

= 2 log10 10

= 2 log 10 10

= 2 …(Simplified)

Question no – (24)

Solution :

(i) Given that log10 2 = x, log10 3 = y

log10 60

= log10 2 + log103 + log1010

= x + y + 1

Question no – (26)

Solution :

(i) log10 x = m + n and, log10y = m – n

= log10 10x/y2 = log10 10n – log10y2

= log10 10 + log 10n – log10 y

= 1 + m + n – 2 (m – n)

= 1 + m + n – 2m – 2n

= 1 – m + 2n

(ii) log x = – 2

= log10 10-2

= – 2 log10 10

= log10 x = log10 1/100

x = 1/100

Question no – (28)

Solution :

Given, x2 + y2 – 2xy = 51xy – 2xy

= (x – y)2 = (7xy)2

= (x – y/7)2 = xy

= x – y/7 = √xy

= log (x – 7/7) log √xy

= log (xy)1/2

= 1/2 log (xy)

= 1/2 [log10x + log10y] …(Proved)

Next Chapter Solution :

Updated: June 20, 2023 — 5:45 am