9th Class Mathematics - Chapter 2

Chapter 2: Logarithms

Exercise 2.2 Solved Notes

A complete solved guide for converting between Logarithmic form and Exponential form, and solving basic logarithmic equations for the unknown variable, completely cleaned of all watermarks and brandings.

Definition of Logarithm Theory

If $a x = y$ (where $a > 0$ , $a \neq 1$ , and $y > 0$ ), then $x$ is called the logarithm of $y$ to the base $a$ . This relation is written as:

log a y = x

Thus, the two forms are completely equivalent:

a x = y \Leftrightarrow log a y = x

Identities to Remember:

$log b 1 = 0$ since $b 0 = 1$
$log b b = 1$ since $b 1 = b$
$log b (1 y) = -log b y$

Question 1 Logarithmic Form

Express each of the following in logarithmic form.

(i) 10³ = 1000

Solution:

By comparing with $a x = y$ , we have base $a = 10$ , exponent $x = 3$ , and result $y = 1000$ .

log 10 1000 = 3

(ii) 2⁸ = 256

Solution:

Here base $a = 2$ , exponent $x = 8$ , and result $y = 256$ .

log 2 256 = 8

(iii) 3^-3 = 127

Solution:

Here base $a = 3$ , exponent $x = -3$ , and result $y = 127$ .

log 3 (127) = -3

(iv) 20² = 400

Solution:

Here base $a = 20$ , exponent $x = 2$ , and result $y = 400$ .

log 20 400 = 2

(v) 16^-1/4 = 12

Solution:

Here base $a = 16$ , exponent $x = - 14$ , and result $y = 12$ .

log 16 (12) = - 14

(vi) 11² = 121

Solution:

Here base $a = 11$ , exponent $x = 2$ , and result $y = 121$ .

log 11 121 = 2

(vii) p = q^r

Solution:

Rearrange as $q r = p$ . Here base $a = q$ , exponent $x = r$ , and result $y = p$ .

log q p = r

(viii) (32)^-1/5 = 12

Solution:

Here base $a = 32$ , exponent $x = - 15$ , and result $y = 12$ .

log 32 (12) = - 15

Question 2 Exponential Form

Express each of the following in exponential form.

(i) log₅ 125 = 3

Solution:

By definition, base $5$ raised to exponent $3$ equals $125$ :

53 = 125

(ii) log₂ 16 = 4

Solution:

Base $2$ raised to exponent $4$ equals $16$ :

24 = 16

(iii) log₂₃ 1 = 0

Solution:

Base $23$ raised to exponent $0$ equals $1$ :

230 = 1

(iv) log₅ 5 = 1

Solution:

Base $5$ raised to exponent $1$ equals $5$ :

51 = 5

(v) log₂ (18) = -3

Solution:

Base $2$ raised to exponent $-3$ equals $18$ :

2 -3 = 18

(vi) 12 = log₉ 3

Solution:

Rearrange as $log 9 3 = 12$ . Base $9$ raised to exponent $12$ equals $3$ :

91/2 = 3   (\text{since } \sqrt{9} = 3)

(vii) 5 = log₁₀ 100,000

Solution:

Rearrange as $log 10 100,000 = 5$ . Base $10$ raised to exponent $5$ equals $100,000$ :

105 = 100,000

(viii) log₄ (116) = -2

Solution:

Base $4$ raised to exponent $-2$ equals $116$ :

4 -2 = 116

Question 3 Solving Logarithmic Equations

Find the value of

x

in each of the following.

(i) log_x 64 = 3

Solution:

1. Convert the logarithmic equation to its equivalent exponential form:

x 3 = 64

2. Express 64 as a perfect cube:

64 = 4 \times 4 \times 4 = 4 3

3. Substitute back into the equation:

x 3 = 4 3

4. Since the exponents are equal, by comparing bases we get:

x = 4

(ii) log₅ 1 = x

Solution:

1. Convert to exponential form:

5 x = 1

2. Any non-zero base raised to power 0 equals 1 ( $50 = 1$ ):

5 x = 5 0

3. By comparing exponents, we obtain:

x = 0

(iii) log_x 8 = 1

Solution:

1. Convert to exponential form:

x 1 = 8

2. Since $x 1 = x$ , we immediately get:

x = 8

(iv) log₁₀ x = -3

Solution:

1. Convert to exponential form:

10 -3 = x

2. Solve the power:

x = 1103 x = 11000 x = 0.001

x = 0.001

(v) log₄ x = 32

Solution:

1. Convert to exponential form:

4 3/2 = x

2. Write 4 as a power of 2 ( $4 = 2 2$ ):

x = (2 2) 3/2

3. Multiply the exponents ( $2 \times 32 = 3$ ):

x = 2 3 x = 2 \times 2 \times 2 = 8

x = 8

(vi) log₂ 1024 = x

Solution:

1. Convert to exponential form:

2 x = 1024

2. Factorize 1024 into base 2 powers:

Base	Division
2	1024
2	512
2	256
2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

So, $1024 = 2 10$ .

3. Substitute into the equation:

2 x = 2 10

4. Comparing the exponents w.r.t base 2, we obtain:

x = 10

Interactive Logarithm Playground Interactive

Select the conversion type and input parameters to format equations dynamically.

Exponential → Logarithmic

Form: b^x = y

Base (b):

Exponent (x):

Result (y):

Logarithmic → Exponential

Form: log_b y = x

Base (b):

Arg (y):

Val (x):

Solve logarithmic equations of the form log_b y = x where one variable is unknown (entered as x).

Log Equation Solver

Leave exactly one input as the letter x to solve for it!

Base (b):

Argument (y):

Value (val):

Test your logarithmic conversion skills with these questions.

1. What is the logarithmic form of 7³ = 343?

(a) log₃ 343 = 7

(b) log₇ 343 = 3 (Correct)

(d) log₇ 3 = 343

Correct Answer: (b) log₇ 343 = 3
Reason: The base is 7, the exponent (log value) is 3, and the argument is 343.

2. Express log₆ (136) = -2 in exponential form.

(a) (-2)⁶ = 36

(b) 6^-2 = 136 (Correct)

(d) 6² = 36

Correct Answer: (b) 6^-2 = 136
Reason: Base 6 raised to exponent -2 equals the argument

136

3. Solve for x: log₃ x = -2.

(a) x = -9

(b) x = 9

(d) x = 0.01

Correct Answer: (c) x = 19
Reason: In exponential form, 3^-2 = x. Thus, x =

132 = 19

4. Solve for x: log_x 27 = 3.

(a) x = 3 (Correct)

(b) x = 9

(d) x = 1

Correct Answer: (a) x = 3
Reason: In exponential form, x³ = 27. Since 27 = 3³, we have x³ = 3³ ⇒ x = 3.

5. Solve for x: log₉ x = 1.5.

(a) x = 13.5

(b) x = 9

(d) x = 81

Correct Answer: (c) x = 27
Reason: In exponential form, 9^1.5 = x. Since 9 = 3², we have x = (3²)^3/2 = 3³ = 27.