Logarithm

Laws of Logarithm Solved Examples on Logarithm Characteristic and Mantissa Properties of Logarithm Properties of Monotonocity of Logarithm Logarithmic Functions Graph Logarithm 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,

\begin{array}{l} \log_{a} b = x \Leftrightarrow a^{x} = b, a > 0, a \neq 1 and b > 0. \end{array}

If a = 10, then we write log b rather than log₁₀b. 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

\begin{array}{l} [since \ln b = \log_{10} b \times \log_{e} 10 = \frac{1}{\log_{10} e} \times \log_{10} b = \frac{1}{0.4343} \log_{10} b = 2.303 \log_{10} b] \end{array}

⇒ Important Points

log 2 = log₁₀2 = 0.3010
log 3 = log₁₀3 = 0.4771
ln 2 = 2.303 log 2 = 0.693
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

\begin{array}{l} a^{\log_{a} b} = b, a > 0, a \neq 1 and b > 0 \end{array}

which is known as the Fundamental Logarithmic Identity.

Corollary 2: The function defined by

\begin{array}{l} f (x) = \log_{a} x, a > 0, a \neq 1 \end{array}

is called logarithmic function. Domain is (0, ∞), and the range is R (set of all real numbers).

Corollary 3: a^x > 0, ∀ x ∈ R

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

Corollary 4:

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

Corollary 5:

$\begin{array}{l} \log_{a} b \to \infty, i f a > 1, b \to \infty \end{array}$
$\begin{array}{l} \log_{a} b \to - \infty, i f 0 < a < 1, b \to \infty \end{array}$

Remarks

‘log’ is the abbreviation of the word ‘logarithm’.
Common logarithm (Brigg’s logarithms). The base is 10.
If x < 0, a > 0 and a ≠ 1, then log_a x is imaginary.
$\begin{array}{l} \log_{a} 1 = 0 (a > 0, a \neq 1) \end{array}$
$\begin{array}{l} \log_{a} a = 1 (a > 0, a \neq 1) \end{array}$
$\begin{array}{l} \log_{(1 / a)} a = - 1 (a > 0, a \neq 1) \end{array}$

\begin{array}{l} If a > 1, \log_{a} x = {\begin{array}{c} + v e, x > 1 \\ 0, x = 1 \\ - v e, 0 < x < 1 \end{array} and if, \end{array}

\begin{array}{l} 0 < a < 1, \log_{a} x = {\begin{array}{c} + v e, 0 < x < 1 \\ 0, x = 1 \\ - v e, x > 1 \end{array} \end{array}

Solved Examples on Logarithm

Example: 1:

\begin{array}{l} Find the value of \log_{\tan 45^{\circ}} \cot 30^{\circ} \end{array}

Solution:

Here, base tan 45⁰ = 1

∴ the log is not defined.

Example: 2:

\begin{array}{l} Find the value of l o g_{(s e c^{2} 60^{0} - t a n^{2} 60^{0})} c o s 60^{0} \end{array}

Solution:

Here, base

\begin{array}{l} = \sec^{2} 60^{\circ} - \tan^{2} 60^{\circ} = 1 \end{array}

∴ the log is not defined.

Example: 3:

\begin{array}{l} Find the value of \log_{(s e c^{2} 30^{\circ} + \cos^{2} 30^{\circ})} 1 \end{array}

Solution:

Since,

\begin{array}{l} \log_{(\sin^{2} 30^{\circ} + \cos^{2} 30^{\circ})} 1 = \log_{1} 1 \neq 1 \end{array}

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

Example: 4:

\begin{array}{l} Find the value of \log_{30} 1 \end{array}

Solution:

\begin{array}{l} \log_{30} 1 = 0 \end{array}

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.

\begin{array}{l} In short, -3 + 0.951823 is written as \overset{―}{3} .9518231 . \end{array}

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

\begin{array}{l} If \log 0.0000279 = \overset{―}{5} .4456042 \end{array}

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

\begin{array}{l} (\overset{―}{4 + 1}), i . e . \overset{―}{5} . \end{array}

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

Log_a (m n) = log_a m + log_a n
In general, log_a(x₁, x₂, x₃,…, x_n) = log_ax₁ + log_ax₂ + log_ax₃ +… + log_ax_n (where x₁, x₂, x₃,…, x_n > 0)
Or

$\begin{array}{l} \log_{a} (\prod_{i = 1}^{n} x_{i}) = \sum_{i = 1}^{n} \log_{a} x_{i}, \forall x_{i} > 0 \end{array}$
where, i = 1, 2, 3, …, n.
$\begin{array}{l} \log_{a} (\frac{m}{n}) = \log_{a} m - \log_{a} n \end{array}$
$\begin{array}{l} \log_{a} m^{α} = α \log_{a} m \end{array}$
$\begin{array}{l} \log_{a^{β}} m = \frac{1}{β} \log_{a} m \end{array}$
$\begin{array}{l} \log_{b} m = \frac{\log_{a} m}{\log_{a} b} \end{array}$
$\begin{array}{l} \log_{b} a . \log_{a} b = 1 \Leftrightarrow \log_{b} a = \frac{1}{\log_{a} b} \end{array}$
$\begin{array}{l} \log_{b} a . \log_{c} b . \log_{a} c = 1 \end{array}$
$\begin{array}{l} \log_{y} x . \log_{z} y . \log_{a} z = \log_{a} x \end{array}$
$\begin{array}{l} e^{\ln a^{x}} = a^{x} \end{array}$

Some Additional Logarithm Properties

Logarithm properties are given below.

$\begin{array}{l} a^{\log_{b} x} = x^{\log_{b} a}, b \neq 1, a, b, x are positive numbers. \end{array}$
$\begin{array}{l} a^{\log_{a} x} = x, a > 0, a \neq 1, x > 0 \end{array}$
$\begin{array}{l} \log_{a^{k}} x = \frac{1}{k} \log_{a} x, a > 0, a \neq 1, x > 0 \end{array}$
$\begin{array}{l} \log_{a} x^{2 k} = 2 k \log_{a} | x |, a > 0, a \neq 1, k \in I \end{array}$
$\begin{array}{l} \log_{a^{^{2 k}}} x = \frac{1}{2 k} \log_{| a |} x, x > 0, a \neq \pm 1 a n d k \in I \tilde{} {10} \end{array}$
$\begin{array}{l} \log_{a^{α}} x^{β} = \frac{β}{α} \log_{a} x, x > 0, a > 0, a \neq 1, α \neq 0 \end{array}$
$\begin{array}{l} \log_{a} x^{2} \neq 2 \log_{a} x, a > 0, a \neq 1 \end{array}$
The domain of log_a (x)² is R ~ {0}, and the domain of log_a x is (0, ∞) are not the same.
$\begin{array}{l} a_{b}^{l o g a} = \sqrt{a}, i f b = a^{2}, a > 0, b > 0, b \neq 1 \end{array}$
$\begin{array}{l} a_{b}^{l o g a} = a^{2}, i f b = \sqrt{a}, a > 0, b > 0, b \neq 1 \end{array}$

Example 1: Solve the equation

\begin{array}{l} 3. x^{\log_{5} 2} + 2^{\log_{5} x} = 64 \end{array}

Solution:

\begin{array}{l} \Rightarrow 3. x^{\log_{5} 2} + 2^{\log_{5} x} = 64 \end{array}

\begin{array}{l} \Rightarrow {3.2}^{\log_{5} x} + 2^{\log_{5} x} = 64 \end{array}

[by extra property (i)]

\begin{array}{l} \Rightarrow {4.2}^{\log_{5} x} = 64 \end{array}

\begin{array}{l} \Rightarrow 2^{\log_{5} x} = 4^{2} = 2^{4} \end{array}

\begin{array}{l} ∴ \log_{5} x = 4 \end{array}

\begin{array}{l} ∴ x = 5^{4} = 625 \end{array}

Example 2:

\begin{array}{l} If 4^{\log_{16} 4} + 9^{\log_{3} 9} = 10^{\log_{x} 83}, find x . \end{array}

Solution:

Since,

\begin{array}{l} 4^{\log_{16} 4} = \sqrt{4} = 2 \end{array}

[by extra property (ix)]

and

\begin{array}{l} 9^{\log_{3} 9} = 9^{2} = 81 \end{array}

[by extra property (viii)]

\begin{array}{l} ∴ 4^{\log_{16} 4} + 9^{\log_{3} 9} = 2 + 81 = 83 = 10^{\log_{x} 83} \end{array}

\begin{array}{l} \Rightarrow \log_{10} 83 = \log_{x} 83 \end{array}

∴ x = 10

Example 3:

\begin{array}{l} Prove that a^{\sqrt{\log_{a} b}} - b^{\sqrt{\log_{b} a}} = 0. \end{array}

Solution:

Since,

\begin{array}{l} a^{\sqrt{(\log_{a} b)}} = a^{\sqrt{\log_{a} b} \times \sqrt{\log_{a} b} \times \sqrt{\log_{b} a}} \end{array}

\begin{array}{l} = a^{\log_{a} b . \sqrt{\log_{b} a}} \end{array}

\begin{array}{l} = b^{\sqrt{\log_{b} a}} \end{array}

[by extra property (ii)]

Hence,

\begin{array}{l} a^{\sqrt{(\log_{a} b)}} - b^{\sqrt{(\log_{b} a)}} = 0 \end{array}

Example 4:

\begin{array}{l} Prove that \frac{\log_{2} 24}{\log_{96} 2} - \frac{\log_{2} 192}{\log_{12} 2} = 3. \end{array}

Solution:

\begin{array}{l} L H S = \frac{\log_{2} 24}{\log_{96} 2} - \frac{\log_{2} 192}{\log_{12} 2} \end{array}

\begin{array}{l} = \log_{2} 24 \times \log_{2} 96 - \log_{2} 192 \times \log_{2} 12 \end{array}

Now, let 12 = λ, then

\begin{array}{l} L H S = \log_{2} 2 λ \times \log_{2} 8 λ - \log_{2} 16 λ \times \log_{2} λ \end{array}

\begin{array}{l} = (\log_{2} 2 + \log_{2} λ) (\log_{2} 8 + \log_{2} λ) - (\log_{2} 16 + \log_{2} λ) \log_{2} λ \end{array}

\begin{array}{l} = (\log_{2} 2 + \log_{2} λ) (\log_{2} 2^{3} + \log_{2} λ) - (\log_{2} 2^{4} + \log_{2} λ) \log_{2} λ \end{array}

\begin{array}{l} = (1 + \log_{2} λ) (3 \log_{2} 2 + \log_{2} λ) - (4 \log_{2} 2 + \log_{2} λ) \log_{2} λ \end{array}

\begin{array}{l} = (1 + \log_{2} λ) (3 + \log_{2} λ) - \log_{2} λ (4 + \log_{2} λ) \end{array}

= 3

= RHS

Properties of Monotonocity of Logarithm

Logarithm with Constant Base

$\begin{array}{l} \log_{a} x > \log_{a} y \Leftrightarrow {\begin{array}{c} x > y > 0, i f a > 1 \\ 0 < x < y, i f 0 < a < 1 \end{array} \end{array}$
$\begin{array}{l} \log_{a} x < \log_{a} y \Leftrightarrow {\begin{array}{c} 0 < x < y, i f a > 1 \\ x > y > 0, i f 0 < a < 1 \end{array} \end{array}$
$\begin{array}{l} \log_{a} x > p \Leftrightarrow {\begin{array}{c} x > a^{p}, i f a > 1 \\ 0 < x < a^{p}, i f 0 < a < 1 \end{array} \end{array}$
$\begin{array}{l} \log_{a} x < p \Leftrightarrow {\begin{array}{c} 0 < x < a^{p}, i f a > 1 \\ x > a^{p}, i f 0 < a < 1 \end{array} \end{array}$

Logarithm with Variable Base

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

Key Points

If a > 1, p > 1, then log_a p > 0
If 0 < a < 1, p > 1, then log_a p < 0
If a > 1, 0 < p < 1, then log_a p < 0
If p > a > 1, then log_a p > 1
If a > p > 1, then 0 < log_a p < 1
If 0 < a < p < 1, then 0 < log_a p < 1
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

Graph of logarithmic function

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

Graph of logarithmic function 2

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

Graph of logarithmic function if x and a are on same side of unity

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

Graph of logarithmic function if x and a are on opposite side of unity

3. Graph of

$\begin{array}{l} y = \log_{a} | x | \end{array}$

Graph of logarithmic function y equals log modulus x

Graphs are symmetrical about Y-axis.

4. Graph of

$\begin{array}{l} y = | \log_{a} | x ‖ \end{array}$

Graph of logarithmic function 4

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

5. Graph of

$\begin{array}{l} | y | = \log_{a} | x ‖ \end{array}$

Graph of logarithmic function 5

6. Graph of

$\begin{array}{l} y = \log_{a} [x], a > 1 a n d x \geq 1 \end{array}$

(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.

Graph of logarithmic function 6

Important Shortcuts to Answer Logarithm Problems

For a non-negative number ‘a’ and
$\begin{array}{l} n \geq 2, n \in N, \sqrt[n]{a} = a^{1 / n} . \end{array}$
The number of positive integers having base a and characteristic n is
$\begin{array}{l} a^{n + 1} - a^{n} . \end{array}$
The logarithm of zero and negative real numbers is not defined.
$\begin{array}{l} | \log_{b} a + \log_{a} b | \geq 2, \forall a > 0, a \neq 1, b > 0, b \neq 1 \end{array}$
$\begin{array}{l} \log_{2} \log_{2} \underset{n t i m e s}{\underset{⏟}{\sqrt{\sqrt{\sqrt{\sqrt{\dots \sqrt{2}}}}}}} = - n \end{array}$
$\begin{array}{l} a^{\sqrt{\log_{a} b}} = b^{\sqrt{\log_{b} a}} \end{array}$
Logarithms to the base 10 are called common logarithms (Brigg’s logarithms).
If
$\begin{array}{l} x = \log_{c} b + \log_{b} c, y = \log_{a} c + \log_{c} a, z = \log_{a} b + \log_{b} a, t h e n x^{2} + y^{2} + z^{2} - 4 = x y z . \end{array}$

Practice Problems on Logarithm

Logarithm examples with solutions are given below.

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

Solution:

\begin{array}{l} \log 10^{11} = 11 \log 10 = 11 \end{array}

and log11¹⁰ = 10log11 = 10 × 1.0414 = 10.414

It is clear that 11 > 10.414

\begin{array}{l} \Rightarrow \log 10^{11} > \log 11^{10} \end{array}

[Since, base = 10]

\begin{array}{l} \Rightarrow 10^{11} > 11^{10} \end{array}

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

\begin{array}{l} \log_{2} (x - 2) < \log_{2^{2}} (x - 2) = \frac{1}{2} \log_{2} (x - 2) \end{array}

\begin{array}{l} \Rightarrow \log_{2} (x - 2) < \frac{1}{2} \log_{2} (x - 2) \end{array}

\begin{array}{l} \Rightarrow \frac{1}{2} \log_{2} (x - 2) < 0 \Rightarrow \log_{2} (x - 2) < 0 \end{array}

\begin{array}{l} \Rightarrow x - 2 < 2^{0} \end{array}

⇒ x – 2 < 1

⇒ x < 3 ……………… (ii)

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

\begin{array}{l} 2 < x < 3 o r x \in (2, 3) \end{array}

Problem 3: If

\begin{array}{l} a^{\log_{b} c} = {3.3}^{\log_{4} 3} {.3}^{\log_{4} 3}^{^{\log_{4} 3}} {.3}^{\log_{4} 3^{^{\log_{4} 3^{\log_{b} c}}}} \dots \infty \end{array}

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

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

Solution:

Option: (c)

\begin{array}{l} a^{\log_{b} c} = 3^{1 + \log_{4} 3} +^{{(\log_{4} 3)}^{2}} +^{{(\log_{4} 3)}^{3}} + \dots \infty \end{array}

\begin{array}{l} = 3^{1 / (1 - \log_{4} 3)} = 3^{1 / \log_{4} (4 / 3)} = 3^{\log_{4 / 3} 4} \end{array}

∴ a = 3, b = 4/3 and c = 4

Hence,

\begin{array}{l} a b c = 3. \frac{4}{3} .4 = 16 \end{array}

Problem 4: Number of real roots of equation

\begin{array}{l} 3^{\log_{3} (x^{2} - 4 x + 3)} = (x - 3) \end{array}

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

Solution: Option (a)

\begin{array}{l} 3^{\log_{3} (x^{2} - 4 x + 3)} = (x - 3) \dots \dots \dots . (i) \end{array}

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

\begin{array}{l} \Rightarrow (x - 1) (x - 3) > 0 \end{array}

⇒ x < 1 or x > 3 ……….(ii)

Equation (i) reduces to

\begin{array}{l} x^{2} - 4 x + 3 = x - 3 \Rightarrow x^{2} - 5 x + 6 = 0 \end{array}

∴ x = 2, 3 ……….(iii)

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

∴ The number of real roots = 0

Problem 5: If

\begin{array}{l} x = \log_{2 a} (\frac{b c d}{2}), y = \log_{3 b} (\frac{a c d}{3}), z = \log_{4 c} (\frac{a b d}{4}), w = \log_{5 d} (\frac{a b c}{5}) \end{array}

\begin{array}{l} and \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} + \frac{1}{w + 1} = \log_{a b c d} N + 1, \end{array}

the value of N is

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

Solution:

Option: (c)

Since,

\begin{array}{l} x = \log_{2 a} (\frac{b c d}{2}) \end{array}

\begin{array}{l} \Rightarrow x + 1 = \log_{2 a} (\frac{2 a b c d}{2}) = \log_{2 a} (a b c d) \end{array}

\begin{array}{l} ∴ \frac{1}{x + 1} = \log_{a b c d} 2 a \end{array}

Similarly,

\begin{array}{l} \frac{1}{y + 1} = \log_{a b c d} 3 b, \frac{1}{z + 1} = \log_{a b c d} 4 c \end{array}

and

\begin{array}{l} \frac{1}{w + 1} = \log_{a b c d} 5 d \end{array}

\begin{array}{l} ∴ \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} + \frac{1}{w + 1} = \log_{a b c d} (2 a \cdot 3 b \cdot 4 c \cdot 5 d) \end{array}

\begin{array}{l} = \log_{a b c d} (120 a b c d) \end{array}

\begin{array}{l} = \log_{a b c d} 120 + 1 \end{array}

\begin{array}{l} = \log_{a b c d} N + 1 \end{array}

[given]

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

(d) None of the above

Solution:

Option: (c)

We have

\begin{array}{l} a = \log_{12} 18 = \frac{\log_{2} 18}{\log_{2} 12} = \frac{1 + 2 \log_{2} 3}{2 + \log_{2} 3} \end{array}

and

\begin{array}{l} b = \log_{24} 54 = \frac{\log_{2} 54}{\log_{2} 24} = \frac{1 + 3 \log_{2} 3}{3 + \log_{2} 3} \end{array}

Putting x = log₂ 3, we have

\begin{array}{l} a b + 5 (a - b) = \frac{1 + 2 x}{2 + x} \cdot \frac{1 + 3 x}{3 + x} + 5 (\frac{1 + 2 x}{2 + x} - \frac{1 + 3 x}{3 + x}) \end{array}

\begin{array}{l} = \frac{6 x^{2} + 5 x + 1 + 5 (- x^{2} + 1)}{(x + 2) (x + 3)} = \frac{x^{2} + 5 x + 6}{(x + 2) (x + 3)} = 1 \end{array}

Problem 7:

\begin{array}{l} The value of \frac{\log_{2} 24}{\log_{96} 2} - \frac{\log_{2} 192}{\log_{12} 2} is \end{array}

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

Solution:

Option: (a)

Set log₂ 12 = a,

\begin{array}{l} \frac{1}{\log_{96} 2} = \log_{2} 96 = \log_{2} (2^{3} \times 12) = 3 + a, \end{array}

\begin{array}{l} \log_{2} 24 = 1 + a, \log_{2} 192 = \log_{2} (16 \times 12) = 4 + a \end{array}

and

\begin{array}{l} \frac{1}{\log_{12} 2} = \log_{2} 12 = a . \end{array}

Therefore, the given expression

\begin{array}{l} = (1 + a) (3 + a) - (4 + a) a = 3 \end{array}

Problem 8: The solution of the equation

\begin{array}{l} 4^{\log_{2} \log x} = \log x - {(\log x)}^{2} + 1 \end{array}

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

Solution:

Option: (c)

log₂ log x is meaningful if x > 1

Since

\begin{array}{l} 4^{\log_{2} \log x} = 2^{2 \log_{2} \log x} = {(2^{\log_{2} \log x})}^{2} = {(\log x)}^{2} \end{array}

\begin{array}{l} [a^{\log_{a} x} = x, a > 0, a \neq 1] \end{array}

So the given equation reduces to

\begin{array}{l} 2 {(\log x)}^{2} - \log x - 1 = 0 \end{array}

\begin{array}{l} \Rightarrow \log x = 1, \log x = - 1 / 2. \end{array}

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, ∞)

(d) (0, 3)

Solution:

log_0.5 (x – 1) < log_0.25 (x – 1)

\begin{array}{l} \Leftrightarrow \log_{0.5} (x - 1) < \log_{{(0.5)}^{2}} (x - 1) \end{array}

\begin{array}{l} \Rightarrow \log_{0.5} (x - 1) < \frac{\log_{0.5} (x - 1)}{\log_{0.5} {(0.5)}^{2}} = \frac{1}{2} \log_{0.5} (x - 1) \end{array}

\begin{array}{l} \Leftrightarrow \log_{0.5} (x - 1) < 0 \end{array}

\begin{array}{l} \Leftrightarrow x - 1 > 1 \Leftrightarrow x \in (2, \infty) \end{array}

Problem 10: If n = 2002 !, evaluate

\begin{array}{l} \frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \frac{1}{\log_{4} n} + \dots + \frac{1}{\log_{2002} n} \end{array}

Solution:

We have,

\begin{array}{l} \frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \frac{1}{\log_{4} n} + \dots + \frac{1}{\log_{2002} n} \end{array}

\begin{array}{l} = \log_{n} 2 + \log_{n} 3 + \log_{n} 4 + \dots + \log_{n} 2002 \end{array}

, Since,

\begin{array}{l} \frac{1}{\log_{b} a} = \log_{a} b \end{array}

\begin{array}{l} = \log_{n} (2.3 .4 \dots .2002) \end{array}

\begin{array}{l} = \log_{n} (2002!) = \log_{n} n = 1 \end{array}

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

\begin{array}{l} \frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y}, \end{array}

prove that x^x y^yz^z = 1.

Solution:

Let

\begin{array}{l} \frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y} = λ \end{array}

\begin{array}{l} \Rightarrow \log x = λ (y - z), \log y = λ (z - x), \log z = λ (x - y) \end{array}

\begin{array}{l} \Rightarrow x \log x + y \log y + z \log z \end{array}

\begin{array}{l} = λ x (y - z) + λ y (z - x) + λ z (x - y) = 0 \end{array}

\begin{array}{l} \Rightarrow \log x^{x} + \log y^{y} + \log z^{z} = 0 \end{array}

\begin{array}{l} \Rightarrow \log (x^{x} y^{y} z^{z}) = 0 \end{array}

\begin{array}{l} \Rightarrow x^{x} y^{y} z^{z} = 1 \end{array}

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

\begin{array}{l} \Rightarrow x > 1 a n d \log_{3} (x - 1) > - \frac{5}{4} \end{array}

\begin{array}{l} \Rightarrow x > 1 a n d x - 1 > 3^{- 5 / 4} \end{array}

\begin{array}{l} \Rightarrow x > 1 a n d x > 1 + \frac{1}{3^{5 / 4}} \end{array}

\begin{array}{l} \Rightarrow x > 1 + \frac{1}{3^{5 / 4}} \dots . . (i) \end{array}

Now,

log₃{5 + 4 log₃ (x – 1)} = 2

\begin{array}{l} \Rightarrow 5 + 4 \log_{3} (x - 1) = 3^{2} \end{array}

\begin{array}{l} \Rightarrow \log_{3} (x - 1) = 1 \end{array}

⇒ 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

\begin{array}{l} 3^{x} - 8 > 0 \Rightarrow 3^{x} > 8 \Rightarrow x > \log_{3} 8 \dots . . (i) \end{array}

Now,

log₃ (3^x – 8) = 2 – x

\begin{array}{l} \Rightarrow 3^{x} - 8 = \frac{3^{2}}{3^{x}} \end{array}

\begin{array}{l} \Rightarrow {(3^{x})}^{2} - 8 (3^{x}) - 9 = 0 \end{array}

\begin{array}{l} \Rightarrow (3^{x} - 9) (3^{x} + 1) = 0 \end{array}

\begin{array}{l} \Rightarrow 3^{x} - 9 = 0 [S i n c e, 3^{x} > 8 T h e r e f o r e, 3^{x} + 1 \neq 0] \end{array}

\begin{array}{l} \Rightarrow 3^{x} = 3^{2} \end{array}

∴ 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

\begin{array}{l} \Rightarrow \log {x^{2 \log x}} = \log (10 x^{2}) \end{array}

\begin{array}{l} \Rightarrow 2 \log x . \log x = \log 10 + \log x^{2} \end{array}

\begin{array}{l} \Rightarrow 2 {(\log x)}^{2} = 1 + 2 \log x \end{array}

\begin{array}{l} \Rightarrow 2 y^{2} - 2 y - 1 = 0, w h e r e y = \log x \end{array}

\begin{array}{l} \Rightarrow y = \frac{2 \pm \sqrt{4 + 8}}{2} \end{array}

\begin{array}{l} \Rightarrow y = 1 \pm \sqrt{3} \end{array}

\begin{array}{l} \Rightarrow \log_{10} x = 1 \pm \sqrt{3} \end{array}

\begin{array}{l} \Rightarrow x = 10^{1 \pm \sqrt{3}} \end{array}

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

\begin{array}{l} \Rightarrow 2^{x} < 9 and x < 3 \end{array}

\begin{array}{l} \Rightarrow x < \log_{2} 9 and x < 3 \end{array}

x < 3 ………. (i)

Now,

log₂ (9-2^x) = 10^{log (3-x)}

\begin{array}{l} \Rightarrow 9 - 2^{x} = 2^{3 - x} \end{array}

\begin{array}{l} \Rightarrow 9 - 2^{x} = \frac{2^{3}}{2^{x}} \end{array}

\begin{array}{l} \Rightarrow 9 (2^{x}) - {(2^{x})}^{2} = 8 \end{array}

\begin{array}{l} \Rightarrow {(2^{x})}^{2} - 9 (2^{x}) + 8 = 0 \end{array}

\begin{array}{l} \Rightarrow y^{2} - 9 y + 8 = 0, w h e r e y = 2^{x} \end{array}

\begin{array}{l} \Rightarrow (y - 8) (y - 1) = 0 \end{array}

⇒ y = 8, 1

\begin{array}{l} \Rightarrow 2^{x} = 8, 1 \end{array}

⇒ x = 3, 0

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

Hence, x = 0

Problem 16: Solve:

\begin{array}{l} {(x + 1)}^{\log (x + 1)} = 100 (x + 1) \end{array}

Solution:

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

Now,

\begin{array}{l} {(x + 1)}^{\log (x + 1)} = 100 (x + 1) \end{array}

\begin{array}{l} \Rightarrow \log {{(x + 1)}^{\log (x + 1)}} = \log {100 (x + 1)} \end{array}

\begin{array}{l} \Rightarrow \log (x + 1) . \log (x + 1) = \log 100 + \log (x + 1) \end{array}

\begin{array}{l} \Rightarrow {\log {(x + 1)}^{2}} = 2 + \log (x + 1) \end{array}

\begin{array}{l} \Rightarrow y^{2} - y - 2 = 0, w h e r e y = \log (x + 1) \end{array}

∴ y = 2, -1

\begin{array}{l} \Rightarrow \log_{10} (x + 1) = 2, \log_{10} (x + 1) = - 1 \end{array}

\begin{array}{l} \Rightarrow x + 1 = 10^{2}, x + 1 = 10^{- 1} \end{array}

⇒ x = 99, x = -0.9

Problem 17: Evaluate

\begin{array}{l} \sqrt[3]{72.3}, i f \log 0.723 = \overset{―}{1} .8591 . \end{array}

Solution:

Let

\begin{array}{l} x = \sqrt[3]{72.3} . \end{array}

Then,

log x = log(72.3)^1/3

\begin{array}{l} \Rightarrow \log x = \frac{1}{3} \log 72.3 \end{array}

\begin{array}{l} \Rightarrow \log x = \frac{1}{3} \times 1.8591 \end{array}

\begin{array}{l} \Rightarrow \log x = 0.6197 \end{array}

\begin{array}{l} \Rightarrow x = a n t i \log (0.6197) \end{array}

⇒ x = 4.166

Problem 18:

\begin{array}{l} Evaluate \sqrt[5]{10076}, if log 100.76 = 2.0029. \end{array}

Solution:

Let

\begin{array}{l} x = \sqrt[5]{10076} . \end{array}

Then,

log x = log (10076)^1/5

\begin{array}{l} \Rightarrow \log x = \frac{1}{5} \log 10076 \end{array}

\begin{array}{l} \Rightarrow \log x = \frac{1}{5} \times 4.0029 \end{array}

\begin{array}{l} \Rightarrow \log x = 0.8058 \end{array}

\begin{array}{l} \Rightarrow x = a n t i \log (0.8058) \end{array}

⇒ x = 6.409

Problem 19: What is logarithm of

\begin{array}{l} 32 \sqrt[5]{4} t o t h e b a s e 2 \sqrt{2} \end{array}

Solution:

Here we can write

\begin{array}{l} 32 \sqrt[5]{4} a s 2^{5} 4^{1 / 5} = {(2)}^{27 / 5} a n d 2 \sqrt{2} a s 2^{\frac{3}{2}} \end{array}

By using the formula

\begin{array}{l} \log_{a} M^{x} = x \log_{a} . M a n d \log_{a^{x}} M = \frac{1}{x} \log_{a} M \end{array}

we can solve it.

\begin{array}{l} \log_{2 \sqrt{2}} 32 \sqrt[5]{4} = \log_{(2^{3 / 2})} (2^{5} 4^{1 / 5}) = \log_{(2^{3 / 2})} {(2)}^{27 / 5} = \frac{2}{3} \frac{27}{5} \log_{2} 2 = \frac{18}{5} = 3.6 \end{array}

Problem 20: Prove that,

\begin{array}{l} \log_{4 / 3} (1. \overset{―}{3}) = 1 \end{array}

Solution:

By solving we get

\begin{array}{l} 1. \overset{―}{3} = \frac{4}{3}, \end{array}

and use the formula

\begin{array}{l} \log_{a} a = 1. \end{array}

\begin{array}{l} \log_{4 / 3} 1. \overset{―}{3} = 1 \end{array}

Let x = 1.333 ….. (i)

10x = 13.3333 ….. (ii)

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

\begin{array}{l} 9 x = 12 \Rightarrow x = 12 / 9, x = 4 / 3; \end{array}

Now

\begin{array}{l} \log_{4 / 3} 1 / \overset{―}{3} = \log_{4 / 3} (4 / 3) = 1 \end{array}

Problem 21: If

\begin{array}{l} N = n! (n \in N, n \geq 2) t h e n lim_{N \to \infty} [{(\log_{2} N)}^{- 1} + {(\log_{3} N)}^{- 1} + \dots + {(\log_{n} N)}^{- 1}] \end{array}

Solution:

Here, by using

\begin{array}{l} \log_{a} b = \frac{1}{\log_{b} a} \end{array}

we can write given expansion as

\begin{array}{l} \log_{N} 2 + \log_{N} 3 + \dots \dots + \log_{N} n \end{array}

and then by using

\begin{array}{l} \log_{a} (M . N) = \log_{a} M + \log_{a} N \end{array}

and N = n!.

\begin{array}{l} \Rightarrow {(\log_{2} N)}^{- 1} + {(\log_{3} N)}^{- 1} + \dots \dots + {(\log_{n} N)}^{- 1} \\ = \log_{N} 2 + \log_{N} 3 + \dots \dots + \log_{N} n = \log_{n} (2.3 \dots . N) = \log_{N} N = 1. \end{array}

Problem 22:

\begin{array}{l} If \log x^{2} - \log 2 x = 3 \log 3 - \log 6 then x equals \end{array}

Solution:

By using

\begin{array}{l} \log_{a} (M . N) = \log_{a} M + \log_{a} N a n d \log_{a} M^{x} = x \log_{a} . M \end{array}

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

\begin{array}{l} 2 \log x - \log 2 - \log x = 3 \log 3 - \log 2 - \log 3 \Rightarrow \log x = 2 \log 3 \Rightarrow x = 9 \end{array}

Problem 23: Prove that,

\begin{array}{l} \log_{2 - \sqrt{3}} (2 + \sqrt{3}) = - 1 \end{array}

Solution:

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

\begin{array}{l} 2 + \sqrt{3} = \frac{1}{2 - \sqrt{3}} . \end{array}

Therefore, by using

\begin{array}{l} \log_{1 / N} N = - 1 \end{array}

we can easily prove this.

\begin{array}{l} \Rightarrow \log_{2 - \sqrt{3}} \frac{1}{2 - \sqrt{3}} \\ = \log_{2 - \sqrt{3}} {(2 - \sqrt{3})}^{- 1} \\ = - 1. \log_{2 - \sqrt{3}} (2 - \sqrt{3}) = - 1 \end{array}

Example 24: Prove that,

\begin{array}{l} \log_{5} \sqrt{5 \sqrt{5 \sqrt{5 \dots \dots . . \infty}}} = 1 \end{array}

Solution:

Here,

\begin{array}{l} \sqrt{5 \sqrt{5 \sqrt{5 \dots \dots . . \infty}}} \end{array}

can be represented as

\begin{array}{l} y = \sqrt{5 y} w h e r e y = \sqrt{5 \sqrt{5 \sqrt{5 \dots \dots . . \infty}}} . \end{array}

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

Let

\begin{array}{l} y = \sqrt{5 \sqrt{5 \sqrt{5 \dots \dots . . \infty}}} \end{array}

\begin{array}{l} \Rightarrow y = \sqrt{5 y} \Rightarrow y^{2} = 5 y o r y^{2} - 5 y = 0 \end{array}

\begin{array}{l} \Rightarrow y (y - 5) = 0 \Rightarrow y = 0, y = 5 \end{array}

\begin{array}{l} ∴ \log_{5} 5 = 1 \end{array}

Problem 25: Prove that,

\begin{array}{l} \log_{2.25} (0. \overset{―}{4}) = - 1 \end{array}

Solution:

Similar to example 3, we can solve it by using

\begin{array}{l} \log_{1 / N} N = - 1. \end{array}

x = 0.4444 ….. (i)

10x = 4.4444 ….. (ii)

Equation (ii) – Equation (i)

\begin{array}{l} 9 x = 4 \Rightarrow x = 4 / 9 \end{array}

Also,

\begin{array}{l} 2.25 = \frac{225}{100} = \frac{9}{4}; \log_{2.25} (0. \overset{―}{4}) = \log_{(\frac{9}{4})} (\frac{4}{9}) = - 1 \end{array}

Problem26: Find the value of

\begin{array}{l} 2^{\log_{6} 18} {.3}^{\log_{6} 3} \end{array}

Solution:

We can solve above problem by using

\begin{array}{l} \log_{a} (M . N) = \log_{a} M + \log_{a} N a n d a^{\log_{e} c} = c^{\log_{e} a} \end{array}

step by step.

\begin{array}{l} \Rightarrow 2^{\log_{6} 18} {(3)}^{\log_{6} 3} = 2^{\log_{6} (6 \times 3)} {.3}^{\log_{6} 3} \\ = 2^{1 + \log_{6} 3} {.3}^{\log_{6} 3} = {2.2}^{\log_{6} 3} {.3}^{\log_{6} 3} (S i n c e, a^{\log_{e} c} = c^{\log_{e} a}) \end{array}

\begin{array}{l} 2. {(3)}^{\log_{6} 2} . {(3)}^{\log_{6} 3} = 2 {(3)}^{\log_{6} 2 + \log_{6} 3} = 2 {(3)}^{\log_{6} (6)} = 2. (3) = 6 \end{array}

Problem 27: Find the value of,

\begin{array}{l} \log_{\sec α} (\cos^{3} α) where α \in (0, π / 2) \end{array}

Solution: Consider

\begin{array}{l} \log_{\sec α} (\cos^{3} α) = x . \end{array}

Therefore, by using formula

\begin{array}{l} y = \log_{a} x \Leftrightarrow a^{y} = x \end{array}

we can write

\begin{array}{l} c o s^{3} α = {(\sec α)}^{x} . \end{array}

Hence, by solving this, we will get the value of x.

Let

\begin{array}{l} \log_{\sec α} \cos^{3} α = x \end{array}

\begin{array}{l} \cos^{3} α = {(\sec α)}^{x} \Rightarrow {(\cos α)}^{3} = {(\frac{1}{\cos α})}^{x} \Rightarrow {(\cos α)}^{3} = {(\cos α)}^{- x} \Rightarrow x = - 3 \end{array}

Problem 28: If k ∈ N, such that

\begin{array}{l} \log_{2} x + \log_{4} x + \log_{8} x = \log_{k} x a n d \forall x \in R^{'} I f k = {(a)}^{1 / b} \end{array}

then find the value of a + b; a ∈ N, b ∈ N and b is a prime number.

Solution:

By using

\begin{array}{l} \log_{b} a = \frac{\log_{c} a}{\log_{c} b} = \frac{\log a}{\log b} \end{array}

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

\begin{array}{l} k = {(a)}^{1 / b} \end{array}

, we can obtain the value of a + b.

Given,

\begin{array}{l} \frac{\log x}{\log 2} + \frac{\log x}{2 \log 2} + \frac{\log x}{3 \log 2} = \frac{\log x}{\log k} \\ \Rightarrow \frac{\log x}{\log 2} [\frac{1}{1} + \frac{1}{2} + \frac{1}{3}] = \frac{\log x}{\log 2} (\frac{11}{6}) = \frac{\log x}{\log k} \\ \Rightarrow \log x [\frac{11}{6} \frac{1}{\log 2} - \frac{1}{\log k}] = 0 \end{array}

Also,

\begin{array}{l} \frac{11}{6} \frac{1}{\log 2} - \frac{1}{\log k} = 0 \Rightarrow \frac{11}{6} = \frac{\log 2}{\log k} \Rightarrow \frac{11}{6} = \log_{k} 2 \end{array}

\begin{array}{l} 2 = k^{\frac{11}{6}}; 2^{6 / 11} = k \Rightarrow {(2^{6})}^{\frac{1}{11}} = k \Rightarrow {(64)}^{\frac{1}{11}} = k \end{array}

\begin{array}{l} Comparing by k = {(a)}^{1 / b}, \end{array}

a = 64, b = 11

a + b = 64 + 11 = 75

Problem 29:

\begin{array}{l} \log_{e} [{(1 + x)}^{1 + x} {(1 - x)}^{1 - x}] = \end{array}

Solution:

\begin{array}{l} \log_{e} {{(1 + x)}^{1 + x} {(1 - x)}^{1 - x}} \\ = (1 + x) \log_{e} (1 + x) + (1 - x) \log_{e} (1 - x) \\ = (1 + x) {x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots \dots} + (1 - x) {- x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - \dots \dots .} \\ = 2 {- \frac{x^{2}}{2} - \frac{x^{4}}{4} - \frac{x^{6}}{6} - \dots . .} + 2 {x^{2} + \frac{x^{4}}{3} + \frac{x^{6}}{5} + \dots \dots} \\ = 2 [x^{2} (1 - \frac{1}{2}) + x^{4} (\frac{1}{3} - \frac{1}{4}) + x^{6} (\frac{1}{5} - \frac{1}{6}) + \dots \dots] \\ = 2 [\frac{x^{2}}{1.2} + \frac{x^{4}}{3.4} + \frac{x^{6}}{5.6} + \dots \dots .] \end{array}

Problem 30: In the expansion of

\begin{array}{l} 2 \log_{e} x - \log_{e} (x + 1) - \log_{e} (x - 1) \end{array}

, the coefficient of x^-4 is

Solution:

\begin{array}{l} 2 \log_{e} x - \log_{e} {(1 + \frac{1}{x}) x} - \log_{e} {(1 - \frac{1}{x}) x} \\ = 2 \log_{e} x - {\log_{e} (1 + \frac{1}{x}) + \log_{e} x} - {\log_{e} (1 - \frac{1}{x}) + \log_{e} x} \\ = - {\log_{e} (1 + \frac{1}{x}) + \log_{e} (1 - \frac{1}{x})} \\ = 2 {\frac{1}{2 x^{2}} + \frac{1}{4 x^{4}} + \dots \dots .} \\ The coefficient of x^{- 4} = 2. \frac{1}{4} = \frac{1}{2} \end{array}

Problem 31:

\begin{array}{l} \log_{e} 2 + \log_{e} (1 + \frac{1}{2}) + \log_{e} (1 + \frac{1}{3}) + \dots . + \log_{e} (1 + \frac{1}{n - 1}) = \end{array}

Solution:

\begin{array}{l} \log_{e} 2 + \log_{e} (\frac{3}{2}) + \log_{e} (\frac{4}{3}) + \dots . + \log_{e} (\frac{n}{n - 1}) \\ = \log_{e} 2 + \log_{e} 3 - \log_{e} 2 + \log_{e} 4 - \log_{e} 3 + \dots \dots + \log_{e} (n) - \log_{e} (n - 1) \\ = \log_{e} n . \end{array}

Problem 32: The coefficient of xⁿ in the expansion of

\begin{array}{l} \log_{e} (1 + 3 x + 2 x^{2}) \end{array}

Solution:

\begin{array}{l} \log (1 + 3 x + 2 x^{2}) = \log (1 + x) + \log (1 + 2 x) \\ = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{x^{n}}{n} + \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{{(2 x)}^{n}}{n} \\ = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} (\frac{1}{n} + \frac{2^{n}}{n}) x^{n} \\ = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} (\frac{1 + 2^{n}}{n}) x^{n} \\ So, coefficient of x^{n} = {(- 1)}^{n - 1} (\frac{2^{n} + 1}{n}) \end{array}

\begin{array}{l} [{(- 1)}^{n} = {(- 1)}^{n + 2} = \dots] \end{array}

Problem 33:

\begin{array}{l} 1 + (\frac{1}{2} + \frac{1}{3}) \frac{1}{4} + (\frac{1}{4} + \frac{1}{5}) \frac{1}{4^{2}} + (\frac{1}{6} + \frac{1}{7}) \frac{1}{4^{3}} + \dots . \infty = \end{array}

Solution:

\begin{array}{l} S = {1 + \frac{{(\frac{1}{2})}^{2}}{2} + \frac{{(\frac{1}{2})}^{4}}{4} + \dots . .} + 2 {\frac{1}{2} + \frac{{(\frac{1}{2})}^{3}}{3} + \frac{{(\frac{1}{2})}^{5}}{5} + \dots . .} - 1 \\ = 1 - \frac{1}{2} \log_{e} (1 + \frac{1}{2}) (1 - \frac{1}{2}) + \log_{e} (\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}) - 1 \\ = - \frac{1}{2} \log_{e} \frac{3}{4} + \log_{e} 3 \\ = \log_{e} 2 \sqrt{3} . \end{array}

Problem 34:

\begin{array}{l} 2 \log x - \log (x + 1) - \log (x - 1) is equal to \\ (1) x^{2} + \frac{1}{2} x^{4} + \frac{1}{3} x^{6} + \dots \dots \infty \\ (2) \frac{1}{x^{2}} + \frac{1}{2 x^{4}} + \frac{1}{3 x^{6}} + \dots \dots \infty \\ (3) - (\frac{1}{x^{2}} + \frac{1}{2 x^{4}} + \frac{1}{3 x^{6}} + \dots \dots \infty) \\ (4) None of these \\ Solution : \\ \begin{matrix} 2 \log x - \log (x + 1) - \log (x - 1) = \log x^{2} - [\log (x + 1) + \log (x - 1)] \\ = \log x^{2} - \log {(x + 1) (x - 1)} \\ = \log x^{2} - \log (x^{2} - 1) = \log \frac{x^{2}}{x^{2} - 1} \\ = - \log (\frac{x^{2} - 1}{x^{2}}) = - \log (1 - \frac{1}{x^{2}}) \\ = - [\frac{- 1}{x^{2}} - \frac{1}{2} {(\frac{1}{x^{2}})}^{2} - \frac{1}{3} {(\frac{1}{x^{2}})}^{3} - \frac{1}{4} {(\frac{1}{x^{2}})}^{4} \dots \dots \dots \infty \\ = \frac{1}{x^{2}} + \frac{1}{2 x^{4}} + \frac{1}{3 x^{6}} + \frac{1}{4 x^{8}} + \dots \dots \dots \infty \end{matrix} \\ A n s w e r : [2] \end{array}

Problem 35:

\begin{array}{l} The coefficient of n^{- r} in the expansion of \log_{10} (\frac{n}{n - 1}) \\ (1) \frac{1}{r \log_{e} 10} \\ (2) \frac{\log_{e} 10}{r} \\ (3) - \frac{\log_{e} 10}{r} \\ (4) None of these \\ Solution : \\ \log_{10} (\frac{n}{n - 1}) \Rightarrow \log_{e} (\frac{n}{n - 1}) \cdot \log_{10} e \\ \begin{matrix} \Rightarrow - \log (\frac{n - 1}{n}) \log_{10} e \\ \Rightarrow - \log (1 - \frac{1}{n}) \log_{10} e \\ = [\frac{1}{n} + \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} + \dots \dots + \frac{1}{m^{r}} + \dots \dots \dots \infty] \log_{10} e \end{matrix} \\ ∴ The coefficient of n^{- r} = \frac{1}{r} \log_{10} e \\ = \frac{1}{r \log_{e} 10} \\ A n s w e r : [1] \end{array}

Problem 36:

\begin{array}{l} \log \frac{(1 + x)^{(1 - x) / 2}}{(1 - x)^{(1 + x) / 2}} is equal to \\ (1) x + \frac{5 x^{3}}{2.3} + \frac{9 x^{5}}{4.5} + \frac{13 x^{7}}{6.7} + \dots \dots . . + \infty \\ (2) x - \frac{5 x^{3}}{2.3} + \frac{9 x^{5}}{4.5} - \frac{13 x^{7}}{6.7} + \dots \dots . . + \infty \\ (3) x - \frac{5 x^{3}}{2.3} - \frac{9 x^{5}}{4.5} - \frac{13 x^{7}}{6.7} - \dots \dots . . - \infty \\ (4) None of these \\ Solution : \log \frac{(1 + x)^{(1 - x) / 2}}{(1 - x)^{(1 + x) / 2}} \\ \begin{matrix} = \frac{1}{2} (1 - x) \log (1 + x) - \frac{1}{2} (1 + x) \log (1 - x) \\ = \frac{1}{2} [\log (1 + x) - \log (1 - x)] - \frac{1}{2} [\log (1 + x) + \log (1 - x)] \\ = \frac{1}{2} \cdot 2 [[x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + \dots \dots .] - \frac{1}{2} \cdot x (- 2) [\frac{1}{2} x^{2} + \frac{x^{4}}{4} + \dots . .] \\ = x + (\frac{1}{3} + \frac{1}{2}) x^{2} + (\frac{1}{5} + \frac{1}{4}) x^{5} + (\frac{1}{7} + \frac{1}{6}) x^{7} + \dots \dots . \\ = x + \frac{5 x^{3}}{2.3} + \frac{9 x^{5}}{4.5} + \frac{13 x^{7}}{6.7} + \dots \dots \dots . \end{matrix} \\ A n s w e r : [1] \end{array}

Logarithms – Video Lesson 1

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Logarithms – Video Lesson 2

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Frequently Asked Questions

Who invented logarithms?

John Napier invented logarithms.

Give the product rule of logarithms.

Product rule: log_a(mn) = log_am + log_an.

Give the quotient rule of logarithms.

Quotient rule: log_a(m/n) = log_am – log_an.

Give two applications of logarithms.

Logarithms are used by biologists to find out the population growth rates.
It is also used to measure the magnitude of earthquakes.

State the power rule of the logarithm.

The logarithm of a quantity in exponential form is equal to the product of the exponent and the logarithm of the base of the exponential term, i.e., log (x^a) = a log x.

Introduction to Logarithm

Laws of Logarithm

Solved Examples on Logarithm

Characteristic and Mantissa

Important Conclusions on Characteristic and Mantissa

Principle Properties of Logarithm

Some Additional Logarithm Properties

Properties of Monotonocity of Logarithm

Logarithm with Constant Base

Logarithm with Variable Base

Key Points

Graphs of Logarithmic Functions

Important Shortcuts to Answer Logarithm Problems

Practice Problems on Logarithm

Recommended Videos

Logarithms – Video Lesson 1

Logarithms – Video Lesson 2

Frequently Asked Questions

Who invented logarithms?

Give the product rule of logarithms.

Give the quotient rule of logarithms.

Give two applications of logarithms.

State the power rule of the logarithm.

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