Logarithm: Concept, Formulae & Solved Examples

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Logarithm is used in various aspects of our daily lives. It is the inverse of exponential equations. The logarithm of a given number is the exponent to which another fixed number must be raised to produce that number. The logarithm is of two types:

Common Logarithm: It is defined as the logarithm with base 10.
Natural Logarithm: It is defined as a logarithm with the base ‘e’, where ‘e’ is an irrational number.

Read Also: NCERT Maths Formulas

Some important logarithmic identities are given below:

Particulars	Formula
Product	log_b (xy) = log_bx + log_by
Quotient	log_b(x/y) = log_bx - log_by
Power	log_b(xp) = p log_b x
Root	log_b(px) = log_b x / p

Read More: Class 12 Maths NCERT Solutions

Sample Questions

Ques. Solve the following logarithmic equation:

log_a x + log_a (x+3) log_a10 (3 Marks)

Ans. log_a [x (x-3)] = log_a10

x (x-3) = 10

x² - 3x - 10 = 0

By splitting the middle term, we get

(x-5) (x+2) = 0

x = 5, x ≠ -2

Hence, x = 5

Ques. Given that p = log_a 4 and q = log_a 5, express each of the following logarithms in terms of p and q. (5 Marks)

log_a 100
log_a 0.4

Ans.

log_a 100 = log_a (25×4)

= log_a 25 + log_a4

= log_a 5² + log_a 4

= 2 log_a 5 + log_a 4

= 2q + p, (p = log_a 4 and q = log_a 5)

log_a0.4 = log_a^2\5

= log_a 2 - log_a 5

= log_a 4^1/2 - log_a 5

= ½ log_a 4 - log_a 5

= ½ p - q

Ques. Given that p = log₂ 3 and q = log₂ 5, express each of the following logarithms in terms of p and q. (5 Marks)

log₂ 45
log₂ 0.3

Ans.

log₂ 45 = log₂ (5×9)

= log₂ 5 + log₂ 9

= log₂ 5 + log₂ 3²

= log₂ 5 + 2 log₂ 3

= q + 2p

log₂ 0.3 = log₂ 3/10

= log₂ 3 - log₂ 10

= log₂ 3 - log₂ (5 × 2)

= log₂ 3 - [log₂ 5 + log₂ 2]

= p - [q + 1]

= p - q - 1

Ques. Find the value of x in:

7^x = 10
log₂x = 9/log₂x (5 Marks)

Ans.

7^x = 10

log 7^x = log 10

x log 7 = 1

x = 1 / log7

x ≈ 1.18

log₂x = 9 / log₂x

= (log₂ x)² = 9

= log₂ x = 3 0r -3

= log₂ x = 3 log₂ 2 or - 3 log₂ 2

= log₂ x = log₂ 8 or log₂ ^1/8

= x = 8 0r ^1/8

Ques. Find the value of x in log_a (x²-10) - log_a x = 2log_a3. (3 Marks)

Ans. log_a (x²-10) - log_a x = 2 log_a 3

log_a [ (x²-10) / x) ] = log_a 3²

log_a [ (x²-10) / x) ] = log_a 9

(x²-10) / x) = 9

x² - 10 = 9x

x²- 9x - 10 = 0

(x + 1) (x - 10) = 0

x ≠ -1, x = 10

Ques. Solve the following logarithmic equation:

log₂ (2x+1) = 2+ log₂x (3 Marks)

Ans. log₂ (2x + 1) = 2 + log₂ x

log₂ (2x + 1) = 2 log₂ 2 + log₂ x

log₂ (2x + 1) = log₂ 4 + log₂ x

log₂ (2x + 1) = log₂ (4x)

2x + 1 = 4x

2x = 1

x = ½

Ques. Solve the following equation and find the value of x:

2 log_a x - log_a (5x - 24) = log_a4 (3 Marks)

Ans. 2 log_a x - log_a (5x - 24) = log_a 4

log_a x² - log_a (5x - 24) = log_a 4

log_a [x² / (5x - 24)] = log_a 4

x² / (5x - 24) = 4

x² = 20x - 96

x² - 20x + 96 = 0

(x-8) (x-12) = 0

x = 8, x = 12

Ques. It is given that x satisfies the logarithmic equation log_a x = 2 (log_ak − log_a2), where k > 0 , a > 0 , a ≠ 1.

Find x in terms of k. Suppose instead that x satisfies log_x (5y+1) = 4 + log_x 3 where x > 0 , x ≠ 1 and y > 0, y ≠ 1.
Solve the above equation expressing y in terms of x (5 Marks)

Ans.

log_a x = 2 (log_a k − log_a 2)

log_a x = 2log_a k − 2log_a 2

log_ax = log_a k² − log_a 4

log_a x = log_a (k²/4)

x = k²/4

log_x (5y + 1) = 4 + log_x 3

log_x (5y + 1) = 4log_x x + log_x 3

log_x (5y + 1) = log_x x⁴ + log_x 3

log_x (5y + 1) = log_x (3x⁴)

5y + 1 = 3x⁴

y = (3x⁴ -1) / 5

Ques. Find the value of y if log (y-1) + log (y + 1) = log₂ 1 (3 Marks)

Ans. log (y - 1) + log (y + 1) = log₂ 1

= log (y - 1) + log (y + 1) = 0

= log [ (y-1) (y+1) ] = 0

Since, log 1 = 0

= (y-1) (y+1) = 1

= y² - 1 = 1

= y² = 2

y = ± √2

Since, the log of negative numbers is not defined.

Therefore, y=√2

Ques. Solve each of the following logarithmic equations. (5 Marks)

log_x16 = log_x9 + 2
log_y27 = 3 + log_y8

Ans.

log_x 16 = log_x 9 + 2

log_x 16 = log_x 9 + 2log_x x

log_x 16 = log_x 9 + log_x x²

log_x 16 = log_x (9x²)

16 = 9x²

16/9 = x²

x = ± 4/3

log_y 27 = 3 + log_y 8

log_y 27 = 3log_y y + log_y 8

log_y 27 = log_y y³ + log_y 8

log_y 27 = log_y (8y³)

log_y 33 = log_y (2y)³

2y = 3

y = 3/2

Ques. Find the value of x in 2log₃ x = 1 + log₃ 7x. (3 Marks)

Ans. 2log₃ x = 1 + log₃ 7x

log₃ x² = log₃ 3 + log₃ 7x

log₃ x² = log₃ 21x

x² = 21x

x² - 21x = 0

x (x - 21) = 0

x = 0, (x - 21) = 0

x - 21 = 0

x = 21

Ques. Solve the following logarithmic expression:

log₄x = log₃9 (3 Marks)

Ans. log₄ x = log₃ 9

log₄ x = log₃ 3²

log₄ x = 2log₃ 3

log₄ x = 2log₄ 4

log₄ x = log₄ 4²

x = 4²

x = 16

Ques. Given that a ≠ 0 , b ≠ 0 , y ≠ 0

2 + log_ab + 3log_ay = 2 log_a(a²y)

Express y in terms of a and b. The expression should be logarithm free. (3 Marks)

Ans. 2 + log_a b + 3log_a y = 2 log_a (a² y)

2log_a a + log_a b + log_a y³ = log_a (a² y)²

log_a a² + log_a b + log_a y³ = log_a (a⁴ y²)

log_a [a² by³] = log_a [a⁴ y²]

a²by³ = a⁴y²

by = a²

y = a²/b

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Maths Formulas and Solved Examples	Bayes’ Theorem	Applications of Integrals

Logarithm: Concept, Formulae & Solved Examples

Sample Questions

CBSE CLASS XII Related Questions

1.
The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :

2.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

3.
Find the value of \(x\) if \(A = B\), where
\[ A = \begin{bmatrix} 6 & 3x \\ 8 & -6 \end{bmatrix} \text{ and } B = \begin{bmatrix} 6 & x^2+2 \\ 8 & x^2-5x \end{bmatrix}. \]

4.
Find the sum of maximum and minimum values of \(f(x) = \sin x\,(1 + \cos x)\) in the interval \([0, \pi]\).

5.
Solve the differential equation \( y + \frac{d}{dx}(xy) = x(\sin x + \log x) \).

6.
For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :
Minimize \(Z = 18x + 10y\)
subject to constraints
\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

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Logarithm: Concept, Formulae & Solved Examples

Sample Questions

CBSE CLASS XII Related Questions

1.The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :

2.Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

3.Find the value of \(x\) if \(A = B\), where\[ A = \begin{bmatrix} 6 & 3x \\ 8 & -6 \end{bmatrix} \text{ and } B = \begin{bmatrix} 6 & x^2+2 \\ 8 & x^2-5x \end{bmatrix}. \]

4.Find the sum of maximum and minimum values of \(f(x) = \sin x\,(1 + \cos x)\) in the interval \([0, \pi]\).

5.Solve the differential equation \( y + \frac{d}{dx}(xy) = x(\sin x + \log x) \).

6.For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :Minimize \(Z = 18x + 10y\)subject to constraints\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

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1.
The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :

2.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

3.
Find the value of \(x\) if \(A = B\), where
\[ A = \begin{bmatrix} 6 & 3x \\ 8 & -6 \end{bmatrix} \text{ and } B = \begin{bmatrix} 6 & x^2+2 \\ 8 & x^2-5x \end{bmatrix}. \]

4.
Find the sum of maximum and minimum values of \(f(x) = \sin x\,(1 + \cos x)\) in the interval \([0, \pi]\).

5.
Solve the differential equation \( y + \frac{d}{dx}(xy) = x(\sin x + \log x) \).

6.
For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :
Minimize \(Z = 18x + 10y\)
subject to constraints
\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]