Logarithm

Laws of LogarithmSolved Examples on LogarithmCharacteristic and MantissaProperties of Logarithm Properties of Monotonocity of LogarithmLogarithmic Functions GraphLogarithm Problems Asked in Exams

Introduction to Logarithm

The logarithm of any positive number, whose base is a number, which is greater than zero and not equal to one, is the index or the power to which the base must be raised in order to obtain the given number.

Mathematically, if a^x = b (where a > 0, ≠ 1), then x is called the logarithm of b to the base a, and we write log_a b = x, clearly b > 0. Thus,

If a = 10, then we write log b rather than log₁₀ b. If a = e, we write ln b rather than log_e b. Here, ‘e’ is called Napier’s base and has a numerical value equal to 2.7182. Also, log₁₀ e is known as the Napierian constant.

i.e., log₁₀ e = 0.4343

∴ ln b = 2.303 log₁₀ b

⇒ Important Points

1. log 2 = log₁₀ 2 = 0.3010
2. log 3 = log₁₀ 3 = 0.4771
3. ln 2 = 2.303 log 2 = 0.693
4. ln 10 = 2.303

Laws of Logarithm

Corollary 1: From the definition of the logarithm of the number b to the base a, we have the identity

which is known as the Fundamental Logarithmic Identity.

Corollary 2: The function defined by

Corollary 3: a^x > 0, ∀ x ∈ R

1. If a > 1, then a^x is monotonically increasing. For example,
   \(\begin{array}{l}{{5}^{2.7}}>{{5}^{2.5}},{{3}^{222}}>{{3}^{111}}\end{array} \)
2. If 0 < a < 1, then a^x is monotonically decreasing. For example,
   \(\begin{array}{l}{{\left( \frac{1}{5} \right)}^{2.7}}<{{\left( \frac{1}{5} \right)}^{2.5}},{{\left( 0.7 \right)}^{222}}<{{\left( 0.7 \right)}^{212}}\end{array} \)

Corollary 4:

1. If a > 1, then a^–∞ = 0. i.e., log_a 0 = –∞ (if a > 1) 
2. If 0 < a < 1, then a^∞ = 0. i.e., log_a 0 = +∞ (if 0 < a < 1) 

Corollary 5:

• \(\begin{array}{l}{{\log }_{a}}b\to \infty ,if\,a>1,b\to \infty\end{array} \)
• \(\begin{array}{l}{{\log }_{a}}b\to -\infty ,if\,0<a<1,b\to \infty\end{array} \)

Remarks

1. ‘log’ is the abbreviation of the word ‘logarithm’.
2. Common logarithm (Brigg’s logarithms). The base is 10.
3. If x < 0, a > 0 and a ≠ 1, then log_a x is imaginary.
4. \(\begin{array}{l}{{\log }_{a}}1=0\left( a>0,a\ne 1 \right)\end{array} \)
5. \(\begin{array}{l}{{\log }_{a}}a=1\left( a>0,a\ne 1 \right)\end{array} \)
6. \(\begin{array}{l}{{\log }_{\left( 1/a \right)}}a=-1\left( a>0,a\ne 1 \right)\end{array} \)

Solved Examples on Logarithm

Example: 1:

Solution:

Here, base tan 45⁰ = 1

∴ the log is not defined.

Example: 2:

Solution:

Here, base

∴ the log is not defined.

Example: 3:

Solution:

Since,

Here, base = 1, ∴ the log is not defined.

Example: 4:

Solution:

Characteristic and Mantissa

The integral part of a logarithm is called the characteristic, and the fractional part (decimal part) is called the mantissa.

i.e., log N = Integer + Fractional or decimal part (+ve)

⇒ The mantissa of the log of a number is always kept positive.

i.e., if log564 = 2.751279, then 2 is the characteristic and 0.751279 is the mantissa of the given number 564.

And if log0.00895 = -2.0481769 = -2 – 0.0481769 = (-2 – 1) + (1 – 0.0481769) = -3 + 0.9518231

Hence, -3 is the characteristic and 0.9518231

(Not 0.0481769) is mantissa of log 0.00895.

Important Conclusions on Characteristic and Mantissa

• If the characteristics of log N be n, then the number of digits in N is (n+1) (Here, N > 1).
• If the characteristics of log N be –n, then there exists (n-1) number of zeroes after the decimal part of N (here, 0 < N < 1).
• If N > 1, the characteristic of log N will be less than the number of digits in an integral part of N.
• If 0 < N < 1, the characteristic of log N is negative, and numerically, it is one greater than the number of zeroes immediately after the decimal part in N.

For example:

1. If log 235.68 = 2.3723227. Here, N = 235.68

∴ Number of digits in an integral part of N = 3

⇒ Characteristic of log 235.68 = N -1 = 3 – 1 = 2

2.

Here, four zeroes immediately after the decimal point in the number 0.0000279 is

Problem: If log 2 = 0.301 and log 3 = 0.477, find the number of digits in 6²⁰.

Solution: Let P = 6²⁰ = (2×3)²⁰

∴ log P = 20 log (2×3) = 20 {log 2+ log 3}

= 20 {0.301 + 0.477} = 20 × 0.778 = 15.560

Since the characteristic of log P is 15, the number of digits in P will be 15 + 1, i.e., 16.

Principle Properties of Logarithm

Following are the logarithm rules:

Let m and n be arbitrary positive numbers, be any real numbers, then

1. Log_a (m n) = log_a m + log_a n
   In general, log_a (x₁, x₂, x₃,…, x_n) = log_a x₁ + log_a x₂ + log_a x₃ +… + log_a x_n (where x₁, x₂, x₃,…, x_n > 0)
   Or
   
   \(\begin{array}{l}{{\log }_{a}}\left( \prod\limits_{i=1}^{n}{{{x}_{i}}} \right)=\sum\limits_{i=1}^{n}{{{\log }_{a}}{{x}_{i}},\forall {{x}_{i}}}>0\end{array} \)
   where, i = 1, 2, 3, …, n.
2. \(\begin{array}{l}{{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n\end{array} \)
3. \(\begin{array}{l}{{\log }_{a}}{{m}^{\alpha }}=\alpha {{\log }_{a}}m\end{array} \)
4. \(\begin{array}{l}{{\log }_{{{a}^{\beta }}}}m=\frac{1}{\beta }{{\log }_{a}}m\end{array} \)
5. \(\begin{array}{l}{{\log }_{b}}m=\frac{{{\log }_{a}}m}{{{\log }_{a}}b}\end{array} \)
6. \(\begin{array}{l}{{\log }_{b}}a.{{\log }_{a}}b=1\,\,\,\,\,\Leftrightarrow \,\,\,\,\,{{\log }_{b}}a=\frac{1}{{{\log }_{a}}b}\end{array} \)
7. \(\begin{array}{l}{{\log }_{b}}a.{{\log }_{c}}b.{{\log }_{a}}c=1\end{array} \)
8. \(\begin{array}{l}{{\log }_{y}}x.{{\log }_{z}}y.{{\log }_{a}}z={{\log }_{a}}x\end{array} \)
9. \(\begin{array}{l}{{e}^{\ln \,{{a}^{x}}}}={{a}^{x}}\end{array} \)

Some Additional Logarithm Properties

Logarithm properties are given below.

1. \(\begin{array}{l}{{a}^{{{\log }_{b}}x}}={{x}^{{{\log }_{b}}a}},b\ne 1,a,b,x\ \text{ are positive numbers.}\end{array} \)
2. \(\begin{array}{l}{{a}^{{{\log }_{a}}x}}=x,a>0,a\ne 1,x>0\end{array} \)
3. \(\begin{array}{l}{{\log }_{{{a}^{k}}}}x=\frac{1}{k}{{\log }_{a}}x,a>0,a\ne 1,x>0\end{array} \)
4. \(\begin{array}{l}{{\log }_{a}}{{x}^{2k}}=2k{{\log }_{a}}\left| x \right|,a>0,a\ne 1,k\in I\end{array} \)
5. \(\begin{array}{l}{{\log }_{{{a}^{^{2k}}}}}x=\frac{1}{2k}{{\log }_{\left| a \right|}}x,x>0,a\ne \pm 1\,\,and\,\,k\in I\tilde{\ }\left\{ 10 \right\}\end{array} \)
6. \(\begin{array}{l}{{\log }_{{{a}^{\alpha }}}}{{x}^{\beta }}=\frac{\beta }{\alpha }{{\log }_{a}}x,x>0,a>0,a\ne 1,\alpha \ne 0\end{array} \)
7. \(\begin{array}{l}{{\log }_{a}}{{x}^{2}}\ne 2{{\log }_{a}}x,a>0,a\ne 1\end{array} \)
   The domain of log_a (x)² is R ~ {0}, and the domain of log_a x is (0, ∞) are not the same.
8. \(\begin{array}{l}a_{b}^{log\;a} =\sqrt{a},\,if\,b={a}^{2},a>0,b>0,b\ne 1\end{array} \)
9. \(\begin{array}{l}a_{b}^{log\;a} =a^{2},\,if\,b=\sqrt{a},a>0,b>0,b\ne 1\end{array} \)

Example 1: Solve the equation

Solution:

Example 2:

Solution:

Since,

and

∴ x = 10

Example 3:

Solution:

Since,

Hence,

Example 4:

Solution:

Now, let 12 = λ, then

= 3

= RHS

Properties of Monotonocity of Logarithm

Logarithm with Constant Base

1. \(\begin{array}{l}{{\log }_{a}}x>{{\log }_{a}}y\Leftrightarrow \left\{ \begin{matrix} x>y>0,\,if\,a>1 \\ 0<x<y,\,if\,0<a<1 \\ \end{matrix} \right.\end{array} \)
2. \(\begin{array}{l}{{\log }_{a}}x<{{\log }_{a}}y\Leftrightarrow \left\{ \begin{matrix} 0<x<y,\,if\,a>1 \\ x>y>0,\,if\,0<a<1 \\ \end{matrix} \right.\end{array} \)
3. \(\begin{array}{l}{{\log }_{a}}x>p\Leftrightarrow \left\{ \begin{matrix} x>{{a}^{p}},\,if\,a>1 \\ 0<x<{{a}^{p}},\,if\,0<a<1 \\ \end{matrix} \right.\end{array} \)
4. \(\begin{array}{l}{{\log }_{a}}x<p\Leftrightarrow \left\{ \begin{matrix} 0<x<{{a}^{p}},\,if\,a>1 \\ x>{{a}^{p}},\,if\,0<a<1 \\ \end{matrix} \right.\end{array} \)

Logarithm with Variable Base

1. log_x a is defined, if a > 0, x > 0, x ≠ 1
2. If a > 1, then log_x a is monotonically decreasing in (0, 1) U (1, ∞)
3. If 0 < a < 1, then log_x a is monotonically increasing in (0, 1) U (1, ∞)

Key Points

1. If a > 1, p > 1, then log_a p > 0
2. If 0 < a < 1, p > 1, then log_a p < 0
3. If a > 1, 0 < p < 1, then log_a p < 0
4. If p > a > 1, then log_a p > 1
5. If a > p > 1, then 0 < log_a p < 1
6. If 0 < a < p < 1, then 0 < log_a p < 1
7. If 0 < p < a < 1, then log_a p > 1

Graphs of Logarithmic Functions

1. Graph of y = log_a x, if a > 1 and x > 0

2. Graph of y = log_a x, if 0 < a < 1 and x > 0

If the number x and the base ‘a’ are on the same side of the unity, then the logarithm is positive.

• y = log_a x, a > 1, x > 1
• y = log_a x, 0 < a < 1, 0 < x < 1

If the number x and the base a are on the opposite sides of the unity, then the logarithm is negative.

• y = log_a x, a > 1, 0 < x < 1
• y = log_a x, 0 < a < 1, x > 1

3. Graph of

Graphs are symmetrical about Y-axis.

4. Graph of

Graphs are the same in both cases, i.e., a > 1 and 0 < a < 1.

5. Graph of

6. Graph of

(Where [ . ] denotes the greatest integer function)

Since, when 1 ≤ x < 2, [x] = 1 ⇒ log_a [x] = 0

When 2 ≤ x < 3, [x] = 2 ⇒ log_a [x] = log_a 2

When 3 ≤ x < 4, [x] = 3 ⇒ log_a [x] = log_a 3 and so on.

Important Shortcuts to Answer Logarithm Problems

1. For a non-negative number ‘a’ and
   \(\begin{array}{l}n\ge 2,n\in N,\sqrt[n]{a}={{a}^{1/n}}.\end{array} \)
2. The number of positive integers having base a and characteristic n is
   \(\begin{array}{l}{{a}^{n+1}}-{{a}^{n}}.\end{array} \)
3. The logarithm of zero and negative real numbers is not defined.
4. \(\begin{array}{l}\left| {{\log }_{b}}a+{{\log }_{a}}b \right|\ge 2,\forall a>0,a\ne 1,b>0,b\ne 1\end{array} \)
5. \(\begin{array}{l}{{\log }_{2}}{{\log }_{2}}\underbrace{\sqrt{\sqrt{\sqrt{\sqrt{…\sqrt{2}}}}}}_{n\,\,times}=-n\end{array} \)
6. \(\begin{array}{l}{{a}^{\sqrt{{{\log }_{a}}b}}}={{b}^{\sqrt{{{\log }_{b}}a}}}\end{array} \)
7. Logarithms to the base 10 are called common logarithms (Brigg’s logarithms).
8. If
   \(\begin{array}{l}x={{\log }_{c}}b+{{\log }_{b}}c,y={{\log }_{a}}c+{{\log }_{c}}a,z={{\log }_{a}}b+{{\log }_{b}}a,\,\,then\,\,{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4=xyz.\end{array} \)

Practice Problems on Logarithm

Logarithm examples with solutions are given below.

Problem 1: If log 11 = 1.0414, prove that 10¹¹ > 11¹⁰.

Solution:

and log11¹⁰ = 10log11 = 10 × 1.0414 = 10.414

It is clear that 11 > 10.414

Problem 2: If log₂ (x – 2) < log₄ (x – 2), find the interval in which x lies.

Solution:

Here, x – 2 > 0

⇒ x > 2 ……………… (i)

and

⇒ x – 2 < 1

⇒ x < 3 ……………… (ii)

From equations (i) and (ii), we get

Problem 3: If

where, a, b, c ∈ Q, the value of abc is

(a) 9 (b) 12 (c) 16 (d) 20

Solution:

Option: (c)

∴ a = 3, b = 4/3 and c = 4

Hence,

Problem 4: Number of real roots of equation

(a) 0 (b) 1 (c) 2 (d) infinite

Solution: Option (a)

Equation (i) is defined if x² – 4x + 3 > 0

⇒ x < 1 or x > 3 ……….(ii)

Equation (i) reduces to

∴ x = 2, 3 ……….(iii)

From equations (ii) and (iii), we get x ∈ Φ.

∴ The number of real roots = 0

Problem 5:  If

(a) 40 (b) 80 (c) 120 (d) 160

Solution:

Option: (c)

Since,

Similarly,

and

On comparing, we have  N = 120

Problem 6:  If a = log₁₂ 18, b = log₂₄ 54, then the value of ab +5 (a–b) is

(a) 0

(b) 4

(c) 1

(d) None of the above

Solution:

Option: (c)

We have

Putting x = log₂ 3, we have

Problem 7:

(a) 3 (b) 0 (c) 2 (d) 1

Solution:

Option: (a)

Set log₂ 12 = a,

and

Therefore, the given expression

Problem 8:  The solution of the equation

(a) x = 1 (b) x = 4 (c) x = 3 (d) x = e²

Solution:

Option: (c)

log₂ log x is meaningful if x > 1

Since

So the given equation reduces to

But for x > 1,

log x > 0 so log x = 1 i.e., x = 3

Problem 9:  If log_0.5 (x – 1) < log_0.25 (x – 1), then x lies in the interval.

(a) (2, ∞)

(b) (3, ∞)

(c) (-∞, 0)

(d) (0, 3)

Solution:

log_0.5 (x – 1) < log_0.25 (x – 1)

Problem 10:  If n = 2002 !, evaluate

Solution:

We have,

Problem 11:  If x, y, z > 0 and, such that

Solution:

Let

Problem 12:  Solve: log₃ {5 + 4 log₃ (x – 1)} = 2

Solution:

Clearly, the given equation is meaningful, if x – 1 > 0 and 5 + 4 log₃ (x – 1) > 0

Now,

log₃ {5 + 4 log₃ (x – 1)} = 2

⇒ x – 1 = 3

∴ x=4

Clearly, x = 4 satisfies (i).

Hence, x = 4 is the solution to the given equation.

Problem 13: Solve log₃ (3^x – 8) = 2 – x

Solution: Clearly, the given equation is meaningful, if

Now,

log₃ (3^x – 8) = 2 – x

∴ x = 2

Clearly, 2 > log₃ 8

Hence, x = 2 is the solution of the given equation.

Problem 14: Solve: x^{2 log x} = 10x²

Solution:

Clearly, the given equation is meaningful for x > 0.

Now,

x^{2 log x} = 10x²

Problem 15:  Solve: log₂ (9-2^x) = 10^{log (3-x)}

Solution:

We observe that the two sides of the given equation are meaningful, if

9 – 2^x > 0 and 3 – x > 0

x < 3 ………. (i)

Now,

log₂ (9-2^x) = 10^{log (3-x)}

⇒ y = 8, 1

⇒ x = 3, 0

But, x = 3 does not satisfy (i).

Hence, x = 0

Problem 16:  Solve:

Solution:

The given equation is meaningful for x + 1 > 0 i.e. x > -1

Now,

∴ y = 2, -1

⇒ x = 99, x = -0.9

Problem 17:  Evaluate

Solution:

Let

log x = log(72.3)^1/3

⇒ x = 4.166

Problem 18: 

Solution:

Let

log x =  log (10076)^1/5

⇒ x = 6.409

Problem 19:  What is logarithm of

Solution:

Here we can write

By using the formula

Problem 20:  Prove that,

Solution:

By solving we get

Let x = 1.333 ….. (i)

10x = 13.3333 ….. (ii)

From equations (i) and (ii), we get

So

Now

Problem 21: If

Solution:

Here, by using

Problem 22: 

Solution:

By using

Clearly, x > 0. Then, the given equation can be written as

Problem 23:  Prove that,

Solution:

By multiplying and dividing by 2 + √3 to 2 – √3, we will get

Therefore, by using

Example 24: Prove that,

Solution:

Here,

Hence, by obtaining the value of y, we can prove this.

Let

Problem 25: Prove that,

Solution:

Similar to example 3, we can solve it by using

x = 0.4444 ….. (i)

10x = 4.4444 ….. (ii)

Equation (ii) – Equation (i)

So

Also,

Problem26:  Find the value of

Solution:

We can solve above problem by using

=

Problem 27:  Find the value of,

Solution: Consider

Therefore, by using formula

Let

Problem 28:  If k ∈ N, such that

Solution:

By using

We can obtain the value of k, and then by comparing it to

Given,

Also,

So

a = 64, b = 11

a + b = 64 + 11 = 75

Problem 29:

Solution:

Problem 30: In the expansion of

Solution:

Problem 31:

Solution:

Problem 32: The coefficient of xⁿ in the expansion of

Solution:

Problem 33:

Solution:

Problem 34:

Problem 35:

Problem 36:

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