Regular engagement with OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Ex 7(A) can boost students’ confidence in the subject.

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

Question 1.
Given an equivalent exponential form for each statement :
(i) log2 8 = 3
(ii) log3 81 = 4
(iii) log2 $$\frac { 1 }{ 2 }$$ = – 1
(iv) log5 $$\frac { 1 }{ 25 }$$= – 2
(v) $$\frac { 1 }{ 2 }$$ = log4 2
(vi) $$\frac { 1 }{ 3 }$$ = log27 3
Solution:
(i) log2 8 = 3 ⇒ 2³ = 8
(ii) log3 81 = 4 ⇒ 34 = 81
(iii) log2 $$\frac { 1 }{ 2 }$$ = – 1 ⇒ 2-1 = $$\frac { 1 }{ 2 }$$
(iv) log2 $$\frac { 1 }{ 25 }$$ = – 2 ⇒ 5-2 = $$\frac { 1 }{ 25 }$$
(v) $$\frac { 1 }{ 2 }$$ = log4 2 ⇒ 4$$\frac { 1 }{ 2 }$$ = 2
(vi) $$\frac { 1 }{ 3 }$$ = log27 3 ⇒ 27$$\frac { 1 }{ 3 }$$ = 3

Question 2.
Give an equivalent logarithmic form for each statement :
(i) 16 = 24
(ii) 25 = 5²
(iii) 81 = 34
(iv) 6° = 1
(v) 8$$\frac { 1 }{ 3 }$$ = 2
(vi) $$\frac { 1 }{ 9 }$$ = 3-2
(vii) $$\frac { 1 }{ 32 }$$ = 2-5
(viii) 101.4969 = 31.4
Solution:
(i) 16 = 24 ⇒ log2 16 = 4
(ii) 25 = 5² ⇒ log5 25 = 2
(iii) 81 = 34 ⇒ log3 81 = 4
(iv) 6° = 1 ⇒ log6 1 = 0
(v) 8$$\frac { 1 }{ 3 }$$ ⇒ log8 2 = $$\frac { 1 }{ 3 }$$
(vi) $$\frac { 1 }{ 9 }$$ = 3-2 ⇒ log3 $$\frac { 1 }{ 9 }$$ = – 2
(vii) $$\frac { 1 }{ 32 }$$ = 2-5 ⇒ log3 $$\frac { 1 }{ 32 }$$ = – 5
(viii) 101.4969 = 31.4 ⇒ log10 31.4 = 1.4969

Question 3.
Find the value of each logarithm given below:
(i) log10 1000
(ii) log2 8
(iii) log3 81
(iv) log10 0.1
(v) log10 0.01
(vi) log10 0.0001
(vii) log2 $$\frac { 1 }{ 4 }$$
(viii) log3 $$\frac { 1 }{ 27 }$$
(ix) log3 1
(x) log$$\frac { 1 }{ 2 }$$ $$\frac { 1 }{ 4 }$$
(xi) log27 9
(xii) log$$\frac { 1 }{ 125 }$$ 125
(xiii) log$$\frac { 5 }{ 6 }$$ 1
(xiv) log$$\frac { 1 }{ 3 }$$ 9
(xv) log10 10
Solution:
(i) log10 1000
Let log10 1000 = x, then 10x = 1000
⇒ 10x = (10)³
Comparing, we get
x = 3
∴ log10 1000 = 3

(ii) log2 8
Let log2 8 = x, then 2x = 8
⇒ 2x = 23
Comparing, we get
x = 3
∴ log2 8 = 3

(iii) log3 81
Let log3 81 = x, then 3x = 81 = 34
∴ 3x = 34
Comparing we get,
x = 4
∴ log3 81 = 4

(iv) log10 0.1
Let log10 0.1 = , then 10x = 0.1 = $$\frac { 1 }{ 10 }$$ = 10-1
Comparing we get,
x = – 1
∴ log10 0.1 = – 1

(v) log10 0.01
Let log10 0.01 = x then 10x = 0.01 = $$\frac { 1 }{ 100 }$$
= $$\frac { 1 }{ 10² }$$
∴ 10x = 10-2
Comparing we get,
x = – 2
∴ log100.01 = – 2

(vi) log10 0.0001
Let log10 0.0001 = x, then 10x = 0.0001
= $$\frac{1}{10000}=\frac{1}{10^4}$$
⇒ 10x = 10-4
Comparing we get, x = – 4
∴ log10 0.0001 = – 4

(vii) log2 $$\frac { 1 }{ 4 }$$
Let log2 $$\frac { 1 }{ 4 }$$ = x, then 2x = $$\frac { 1 }{ 4 }$$ = $$\frac { 1 }{ 2² }$$ = 2-2
Comparing we get,
x = – 2
∴ log2 $$\frac { 1 }{ 4 }$$ = – 2

(viii) log3 = $$\frac { 1 }{ 27 }$$
Let log3 $$\frac { 1 }{ 27 }$$ = x, then 3x = $$\frac{1}{27}=\frac{1}{3^3}$$
⇒ 3x = 3-3
Comparing, we get
x = – 3
∴ log3 $$\frac { 1 }{ 27 }$$ = – 3

(ix) log3 1
Let log3 1 = x, then 3x = 1 = 3° (∵ 3° = 1)
Comparing, we get
x = 0
∴ log3 1 = 0

(x) $$\log _{\frac{1}{2}} \frac{1}{4}$$
Let $$\log _{\frac{1}{2}} \frac{1}{4}=x \text {, then }\left(\frac{1}{2}\right)^x=\frac{1}{4}=\left(\frac{1}{2}\right)^2$$
Comparing, we get
x = 2
∴ $$\log _{\frac{1}{2}} \frac{1}{4}$$ = 2

(xi) log27 9
Let log27 9 = x, then 27x = 9
[(3)³]x = (3)² ⇒ 33x = 3²
Comparing, we get
3x = 2 ⇒ x = $$\frac { 2 }{ 3 }$$
∴ log27 9 = $$\frac { 2 }{ 3 }$$

(xii) log$$\frac { 1 }{ 125 }$$ 125
Let log$$\frac { 1 }{ 125 }$$ 125 = x, then $$\left(\frac{1}{5}\right)^x=\frac{1}{125}=\left(\frac{1}{5}\right)^3$$
Comparing, we get
x = 3
∴ log$$\frac { 1 }{ 125 }$$ 125 = 3

(xiii) log$$\frac { 5 }{ 6 }$$ 1

(xiv) log$$\frac { 1 }{ 3 }$$ 9
Let log$$\frac { 1 }{ 3 }$$ 9, then ($$\frac { 1 }{ 3 }$$)x = 9 = 3²
⇒ $$\left(\frac{1}{3}\right)^x=\left(\frac{1}{3}\right)^{-2}$$
Comparing, we get
x = – 2
∴ log$$\frac { 1 }{ 3 }$$ 9 = – 2

(xv) log10 10
Let log10 10 = x, then 10x = 10 = (10)1
Comparing, we get x = 1
∴ log10 10 = 1

Question 4.
Find the value of x
(i) logx 216 = 3
(ii) log4 x = – 4
(iii) log3 x = 0
(iv) log8 x = $$\frac { 2 }{ 3 }$$
(v) log10 100 = x
(vi) log2 0.5 = x
Solution:
(i) logx 216 = 3
⇒ x³ = 216 = (6)³
Comparing, we get
x = 6

(ii) log4 x = – 4
⇒ (4)-4 = x ⇒ $$\frac{1}{4^4}$$ = x
⇒ x = $$\frac { 1 }{ 256 }$$
∴ x = $$\frac { 1 }{ 256 }$$

(iii) log3 x = 0 ⇒ 3° = x
⇒ x = 1
∴ x = 1

(iv) log8 x = $$\frac { 2 }{ 3 }$$ ⇒ $$(8)^{\frac{2}{3}}$$
⇒ x = $$\left(2^3\right)^{\frac{2}{3}}=2^{3 \times \frac{2}{3}}$$ = 2²
⇒ x = 2 x 2 = 4
∴ x = 4

(v) log10 100 = x ⇒ 10x = 100
⇒ 10x = (10)²
Comparing, we get
x = 2

(vi) log2 0.5 = x ⇒ 2x = 0.5 = $$\frac { 1 }{ 2 }$$
⇒ 2x = (2)-1
Comparing both sides
x = – 1

Question 5.
Answer true or false : If log10x = a, then 10a = x.
Solution:
∵ log10 x = a ⇒ 10a = x
Which is true.