Table of Contents

• 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, with a base greater than zero and not equal to one, is the exponent to which the base must be raised to obtain the given number.

Mathematically: If (a^x = b) (where (a > 0, \ne 1)) then (x) is called the logarithm of (b) to the base (a) and we write (\log_a b = x), clearly (b > 0). Thus, (\log_a b = x \Leftrightarrow a^x = b, a > 0, a \ne 1 \ \text{and}\ b > 0.)

If a = 10, then we write log_b rather than log_{10} b. If a = e, we write ln_b rather than log_e b. Here, ’e’ is called as Napier’s base and has numerical value equal to 2.7182. Also, log_{10} e is known as Napierian constant.

log₁₀ e = 0.4343

ln b = 2.303 log_{10} b

[\left[\text{since}\ \ln ,b=2.303\ {{\log }_{10}}b \right]]

Important Points ⇒ ⇒ Key Takeaways

1. log₂ 2 = log₁₀ 2 = 0.3010

2. log₃ = log₁₀ 3 = 0.4771

3. ln 2 = 2.303
log 2 = 0.693

ln 10 = 2.303

Laws of Logarithm

Corollary 1: $\log_a b = \frac{\log_c b}{\log_c a}$

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

Which is known as the Fundamental Logarithmic Identity?

Corollary 2: The function defined by (f\left( x \right)={{\log }_{a}}x,a>0,a\ne 1) is known as a logarithmic function. It has a domain of (0, ∞) and a range of R (the set of all real numbers).

Corollary 3: For all $x \in \mathbb{R}$, $x > 0$

1. If (a > 1), then (ax) is monotonically increasing. For example, (5^{2.7} > 5^{2.5}) and (3^{222} > 3^{111}).

If (0 < a < 1), then (ax) 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 $a^{-\infty} = 0$, i.e. $\log_a 0 = -\infty$ (if $a > 1$).

If $0 < a < 1$, then $a^{\infty} = 0$. i.e. $\log_a 0 = +\infty$ (if $0 < a < 1$)

Corollary 5:

If $a > 1$, then $\log_{a}b \rightarrow \infty$

If 0 < a < 1, then $\log_{a} b \rightarrow -\infty$ and $\log_{a} b \rightarrow \infty$

Remarks: It is important to remember that the work we do is for the benefit of others.

1. log is the abbreviation of the word logarithm.

2. Common logarithm (Brigg’s logarithms): The base is 10.

If x < 0, a > 0 and a ≠ 1, then log_a x is an imaginary number.

4. $$\log_{a} 1 = 0 \quad \text{(for } a > 0, a \neq 1)$$

5. $$\log_a a = 1 \quad \text{(where $a > 0$ and $a \neq 1$)}$$

6. (\log_{\left( -1 \right)} \left( 1/a \right) ) (where (a > 0) and (a \ne 1))

(\begin{array}{l}\text{If}\ a>1,\ x>1 \Rightarrow {{\log }_{a}}x=+ve,\ x=1 \Rightarrow {{\log }_{a}}x=0,\ 0<x<1 \Rightarrow {{\log }_{a}}x=-ve \ \text{and if}\ a\le 1,\end{array})

(\begin{array}{l} 0 < a < 1, \quad {{\log }_a}x = \begin{cases} +ve, \quad 0 < x < 1 \ 0, \quad x = 1 \ -ve, \quad x > 1 \end{cases} \end{array} )

Solved Examples of Logarithms

Example 1: (\begin{array}{l}\text{The value of }\ {{\log }_{\tan 45{}^\circ }}\cot 30{}^\circ \text{ is } \frac{1}{2}\end{array} )

Given:

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Solution:

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base tan 450 = 1

log is not defined.

Example 2: (\begin{array}{l}\text{Find the value of }\ log_{\left(sec^2 60^0 - tan^2 60^0\right)} cos 60^0\end{array} )

Given:

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Solution:

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Here, base $$\sec^2 60^\circ - \tan^2 60^\circ = 1$$

log is not defined.

Example 3: (\log_{\left(sec^2 30^\circ + \cos^2 30^\circ\right)} 1)

Given:

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Solution:

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Since $$\log_{\left(\sin^230^\circ + \cos^230^\circ\right)} 1 = \log_11 \neq 1$$

log is not defined, since base = 1.

Example 4: (\begin{array}{l}\text{Find the value of }\ {\log_{30}}1\end{array} )

Answer: (\begin{array}{l}\text{The value of }\ {\log_{30}}1 = 0 \end{array} )

Given Text:

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Solution:

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‘\(\log_{30} 1 = 0\)’

Characteristic and Mantissa

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

log N = Integer + Fractional/Decimal Part (positive)

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

The given number 564 can be expressed as log564 = 2.751279, where 2 is the characteristic and 0.751279 is the mantissa.

-2.0481769 = log_0.00895 = -2 - 0.0481769 = (-2 - 1) + (1 - 0.0481769) = -3 + 0.9518231

Hence, the characteristic is -3 and the mantissa is 0.9518231.

The mantissa of log 0.00895 is not 0.0481769.

(\begin{array}{l}\text{In short, -3 + 0.951823 is written as}\ \overline{-0.048177}.\end{array} )

Important Conclusions on Characteristic and Mantissa

If the characteristics of log N is n, then the number of digits in N is (n+1) (where N > 1).

If the characteristics of log N be -n, then there exists (n-1) number of zeroes after the decimal part of N, where 0 < N < 1.

If $$N > 1$$, the characteristic of $$\log N$$ will be less than the number of digits in the 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:

This sentence provides an example.

For Example: This sentence provides an example.

1. If $N = 235.68$, then $\log N = 2.3723227$.

The number of digits in the integral part of N is 3.

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

2. (\begin{array}{l}\text{If}\ 10^{\overline{5}.4456042}=0.0000279\end{array} )

The number 0.0000279 has ( \overline{5} ) zeroes immediately after the decimal point, i.e. ( \overline{4+1} ).

**Solution:**The number of digits in 620 is 3.

Solution: P = (2×3)²⁰

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

= 15.560

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

Principle Properties of Logarithms

Following are the rules for logarithms:

Let $m$ and $n$ be arbitrary positive numbers, and let $x$ and $y$ be any real numbers, then

In general, loga (x1, x2, x3,…, xn) = loga x1 + loga x2 + loga x3 +… + loga xn (where x1, x2, x3,…, xn > 0)

Either

$\log_a \left(\prod\limits_{i=1}^{n}{x_i} \right) = \sum\limits_{i=1}^{n}{\log_a x_i}, \quad \forall x_i > 0$

Where, $i = 1, 2, 3, \dots, n$.

2. $\log_a(m/n) = \log_a m - \log_a n$

3. $$\log_{a}m^{\alpha} = \alpha \log_{a}m$$

4. $$\log_{a^\beta}m = \frac{1}{\beta}\log_am$$

$$5. \log_b m = \frac{\log_a m}{\log_a b}$$

6. $$\log_b a \log_a b = 1 \Leftrightarrow \log_b a = \frac{1}{\log_a b}$$

7. (\log_a c = \frac{\log_b a}{\log_b c})

(\log_{a}x = \log_{y}x \cdot \log_{z}y \cdot \log_{a}z)

9. $$e^{\ln a^x} = a^x$$

More Logarithm Properties

Logarithm Properties:

1. $$\begin{array}{l}{a^{\log_b x}} = {x^{\log_b a}}, \quad b \ne 1, \quad a, b, x \text{ are positive numbers.}\end{array}$$

2. $$\begin{array}{l}{a^{\log_a x} = x}, \quad a > 0, \quad a \ne 1, \quad x > 0\end{array}$$

3. $$\log_{a^k} x = \frac{1}{k} \log_a x, \quad a > 0, a \neq 1, x > 0$$

4. $$\log_{a}x^{2k} = 2k\log_{a}\left| x \right|, \quad a > 0, \quad a \neq 1, \quad k \in \mathbb{I}$$

5. $$\log_{a^{2k}} x = \frac{1}{2k} \log_{|a|} x, \quad x > 0, \quad a \neq \pm 1, \quad k \in \mathbb{I}{10}$$

6. $\log_{a^{\alpha}} x^{\beta} = \frac{\beta}{\alpha}\log_a x$, $x > 0$, $a > 0$, $a \neq 1$, $\alpha \neq 0$

7. $\log_{a}{x^2} \neq 2\log_{a}x, a > 0, a \neq 1$

The domain of loga (x)^2 is $\mathbb{R} \setminus {0}$, whereas the domain of loga x is $(0, \infty)$, which are not the same.

8. $$a_{b}^{log;a} = \sqrt{a}, \text{ if } b = a^{2}, a > 0, b > 0, b \ne 1$$

9. $$a_{b}^{log;a} = a^{2}, \text{if } b=\sqrt{a}, a>0, b>0, b\ne 1$$

Example 1: Solve the equation $$3.x^{\log_{5};2} + 2^{\log_{5};x}=64$$

Given:

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Solution:

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

\(\Rightarrow 3.2^{\log\_{5}\;x} + 2^{\log\_{5}\;x}=64\)

By extra property (i)

(\begin{array}{l}\Rightarrow \log_5x = \frac{\log_{4.2} 64}{\log_{4.2} 4.2}\end{array} )

‘(\begin{array}{l}\Rightarrow \log_{5}x=2\log_{5}4=2^2\end{array} )’

$$\therefore \log_{5}x = 4$$

(\therefore x=5^4=625)

Example 2:

If $$4^{\log_{16}4} + 9^{\log_3 9} = 10^{\log_x 83},$$ find x.

Given:

Hello World

Solution:

Hello World

Since, $$4^{\log_{16} 4} = \sqrt{4} = 2$$ [by extra property (ix)]

And (\begin{array}{l}{9}^{{{\log }_{3}}9}=9^2=81\end{array}) [by extra property (viii)]

(\therefore {{10}^{{{\log }_{x}}83}}=83)

‘(\begin{array}{l}\Rightarrow x = 10\end{array} )’

x = 10

Example 3: (\begin{array}{l}\text{Prove that}\ {{a}^{\sqrt{{{\log }_{a}}b}}}-{{b}^{\sqrt{{{\log }_{b}}a}}}=0.\end{array} )

Proof: Let (x = \sqrt{{{\log }_{a}}b}). Then,

[{{a}^{x}}-{{b}^{\frac{1}{x}}}={{a}^{\sqrt{{{\log }_{a}}b}}}-{{b}^{\sqrt{{{\log }_{b}}a}}}=0]

Since (a, b > 0), (x > 0). Thus,

[{{a}^{x}}={{b}^{\frac{1}{x}}}]

Taking the (x^{th}) root of both sides,

[a = b^{\frac{1}{x^2}}]

Rearranging,

[{{\log }_{a}}b={{x}^{2}}]

Substituting back into the original equation,

[{{a}^{\sqrt{{{\log }_{a}}b}}}-{{b}^{\sqrt{{{\log }_{b}}a}}}={{a}^{x}}-{{b}^{\frac{1}{x}}}=0]

Therefore,

[{{a}^{\sqrt{{{\log }_{a}}b}}}-{{b}^{\sqrt{{{\log }_{b}}a}}}=0]

Given:

Welcome to the world of programming!

Solution:

Welcome to the world of programming! :smiley:

Since, $\begin{array}{l},,,,,{{a}^{\sqrt{\left( {{\log }{a}}b \right)}}}={{a}^{\sqrt{{{\log }{a}}b} \times \sqrt{{{\log }{a}}b} \times \sqrt{{{\log }{b}}a}}}\end{array}$

\(\sqrt{a^{\log_a b}\cdot \log_b a}\)

(\begin{array}{l}={{b}^{{a}^{\log_{b}a}}}\end{array} ) [by extra property (ii)]

Hence, $$\left({a}^{\sqrt{\left( {{\log }_{a}}b \right)}}\right) - \left({b}^{\sqrt{\left( {{\log }_{b}}a \right)}}\right) = 0$$

Example 4: (\begin{array}{l}\text{Prove that}\ \frac{{{\log }_{96}}2}{{{\log }_{2}}24}-\frac{{{\log }_{12}}2}{{{\log }_{2}}192}=3.\end{array} )

Given:

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Solution:

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(\begin{array}{l}LHS=\frac{{{\log }_{96}}24}{{{\log }_{2}}2}-\frac{{{\log }_{12}}192}{{{\log }_{2}}2}\end{array} )

(\begin{array}{l}{{\log }_{2}}24\times {{\log }_{2}}96-{{\log }_{2}}192\times {{\log }_{2}}12\end{array})

Now, let $\lambda = 12$, then

(\begin{array}{l}LHS={{\log }_{2}}2\lambda + {{\log }_{2}}8\lambda - {{\log }_{2}}16\lambda - {{\log }_{2}}\lambda\end{array} )

(\begin{array}{l}=\left( {{\log }_{2}}2\cdot {{\log }_{2}}8 \right)-\left( {{\log }_{2}}16 \right){{\log }_{2}}\lambda +\left( {{\log }_{2}}2+{{\log }_{2}}8+{{\log }_{2}}16 \right){{\log }_{2}}\lambda \end{array} )

(\begin{array}{l}=\left( {{\log }_{2}}2+{{\log }_{2}}\lambda \right)\left( {{\log }_{2}}8+{{\log }_{2}}\lambda \right)-\left( {{\log }_{2}}{{16}}+{{\log }_{2}}\lambda \right){{\log }_{2}}\lambda\end{array} )

(\begin{array}{l}=\left( 1+{{\log }_{2}}\lambda \right)\left( 3{{\log }_{2}}2+{{\log }_{2}}\lambda \right)\-\left( 4{{\log }_{2}}2+{{\log }_{2}}\lambda \right){{\log }_{2}}\lambda\end{array} )

(\begin{array}{l}=\left( 1+{{\log }_{2}}\lambda \right)\left( 3+{{\log }_{2}}\lambda \right) - 4{{\log }_{2}}\lambda - {{\log }_{2}}^2\lambda^2 \end{array})

3 =

RHS = Right Hand Side

Properties of Monotonicity of Logarithm

Logarithm with a Constant Base

1. $$\log_a x > \log_a y \Leftrightarrow \begin{cases} x > y > 0, & \text{if } a > 1 \ 0 < x < y, & \text{if } 0 < a < 1 \end{cases}$$

2. $$\log_a x < \log_a y \iff \begin{cases} 0 < x < y, & \text{if } a > 1\ x > y > 0, & \text{if } 0 < a < 1 \end{cases}$$

3. ${\log_a}x > p \Leftrightarrow \begin{cases} x > a^p, & \text{if } a > 1 \ 0 < x < a^p, & \text{if } 0 < a < 1 \end{cases}$

$4. \log_a x < p \Leftrightarrow \begin{cases} 0 < x < a^p, & \text{if } a > 1 \ x > a^p, & \text{if } 0 < a < 1 \end{cases}$

Logarithm with Variable Base

1. Log_x a is defined, if a > 0, x > 0, and x ≠ 1.

2. If a > 1, then $\log_x{a}$ is monotonically decreasing in $(0, 1) \cup (1, \infty)$

If 0 < a < 1, then the logarithm of x with respect to a is monotonically increasing in the intervals (0, 1) and (1, ∞)

Key Points

1. If $$a > 1$$ and $$p > 1$$, then $$\log_a p > 0$$

If $$0 < a < 1$$ and $$p > 1$$, 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 $x > 0$

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, where a > 1 and x > 1

y = log_a x, where 0 < a < 1 and 0 < x < 1

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

y = log_a x, where a > 1 and 0 < x < 1

y = log_a x, where 0 < a < 1 and x > 1

3. Graph of (y=\log_a \left| x \right|)

Graphs are reflectionally symmetric about the Y-axis.

4. Graph of (\left|y = \log_a \left|x\right|\right|)

Graphs are the same in both cases, regardless of whether a > 1 or 0 < a < 1.

5. Graph of (\left| y \right| = \left| \log_a \left| x \right| \right| )

6. Graph of (y = \log_a(x), \text{where } a > 1 \text{ and } x \ge 1)

The greatest integer function of [x] is [ . ]

[ . ] is the greatest integer function of [x].

Since, when $1 \leq x < 2$, $[x] = 1 \implies \log_a[x] = 0$

When $$2 \leq x < 3, x = 2 \Rightarrow \log_a x = \log_a 2$$

When $3 \leq x < 4$, $x = 3 \Rightarrow \log_a x = \log_a 3$ and so on.

Essential Shortcuts for Solving Logarithmic Equations

1. For a non-negative number a and (n \ge 2, n \in \mathbb{N}), (\sqrt[n]{a} = a^{1/n}).

2. The number of positive integers having base a and characteristic n is (a^{n+1}-a^n).

The logarithm of zero and negative real numbers is not defined.

4. $$\left| \log_b a + \log_a b \right| \geq 2, \quad \forall a > 0, a \neq 1, b > 0, b \neq 1$$

5. (\begin{array}{l}{{\log }_{2}}{{\log }_{2}}\underbrace{\sqrt[2^n]{2}}_{n,,times}=-n\end{array} )

6. $$\begin{array}{l}{{a}^{\sqrt{{{\log }_{a}}b}}}={{b}^{\sqrt{{{\log }_{b}}a}}}\end{array}$$

Logarithms to the base 10 are referred to as common logarithms or Brigg’s logarithms.

If $$x = \log_c b + \log_b c, \quad y = \log_a c + \log_c a, \quad z = \log_a b + \log_b a,$$ then $$x^2 + y^2 + z^2 - 4 = xyz.$$

#Practice Problems on Logarithms

Logarithm examples with solutions are provided below.

Problem 1: Prove that $\log_{11}(1011) > \log_{11}(1110)$.

Given

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Solution:

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(\log {{10}^{11}} = 11 \log 10 = 11)

10log₁₁ = 10 × 1.0414 = 10.414

It is clear that 11 is greater than 10.414

(\begin{array}{l}\Rightarrow \log_{10}{10^{11}} > \log_{10}{11^{10}}\end{array} ) [Since, base = 10]

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

Problem 2: Find the interval in which x lies if log2 (x – 2) < log4 (x – 2).

Given:

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Solution:

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x > 2

⇒ x > 2 \\ (i)

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

(\begin{array}{l}\Rightarrow {{\log }_{2}}\left( x-2 \right) \lt \frac{1}{2}{{\log }_{2}}\left( x-2 \right)\end{array} )

(\begin{array}{l}\Rightarrow \frac{1}{2}{{\log }_{2}}\left( x-2 \right)<0 \\ \Downarrow 2{{\log }_{2}}\left( x-2 \right)<0 \\ \Downarrow {{\log }_{2}}\left( x-2 \right)<0\end{array} )

‘(\begin{array}{l}x-2<2\end{array})’

x - 2 < 1

⇒ x < 3 \\ \\ (ii)

From equations (i) and (ii), we obtain:

(\begin{array}{l}x \in \left( 2,3 \right),,or,,2<x<3\end{array})

Problem 3: If $$\begin{array}{l}{{a}^{{{\log }_{b}}c}}={{3.3}^{{{\log }_{4}}3}}{{.3}^{{{\log }_{4}}3}}^{^{{{\log }_{4}}3}}{{.3}^{{{\log }_{4}}{{3}^{^{{{\log }_{4}}{{3}^{{{\log }_{b}}c}}}}}}}…\infty\end{array}$$

Where $a, b, c \in \mathbb{Q}$, the value of $abc$ is $\displaystyle a \cdot b \cdot c$

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

Given:

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Solution:

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Option: (c)

$$\begin{array}{l}{a}^{{{\log }_{b}}c}={{3}^{1+{{\log }_{4}}3}}{{+}^{{{\left( {{\log }_{4}}3 \right)}^{2}}}}{{+}^{{{\left( {{\log }_{4}}3 \right)}^{3}}}}+…\infty\end{array}$$

(\begin{array}{l}={{3}^{{{\log }_{4/3}}4}}={{3}^{{{\log }_{4}}\left( 4/3 \right)}}={{3}^{1/\left( 1-{{\log }_{4}}3 \right)}}\end{array})

a = 3, b = 4/3, c = 4

Hence, $$abc = 3 \times \frac{4}{3} \times 4 = 16$$

Problem 4: Number of real roots of equation $$3^{\log_3(x^2-4x+3)} = (x-3)$$ is

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

Option (a)

(\begin{array}{l}{{\log }_{3}\left( {{x}^{2}}-4x+3 \right)={{3}^{(x-3)}} ……….(i)}\end{array})

Equation (i) is defined if x^2 - 4x + 3 > 0

\((x-1)(x-3) > 0\)

x < 1 **OR** x > 3

Equation (i) reduces to $$\begin{array}{l}{{x}^{2}}-5x+6=0\end{array} \Rightarrow {{x}^{2}}-4x+3=x-3$$

x = 2, 3, …, (iii)

From equations (ii) and (iii), we get $$x \in \Phi$$.

Number of real roots = 0.

Problem 5: If $$\begin{array}{l}x={{\log }_{2a}}\left( \frac{bcd}{2} \right),y={{\log }_{3b}}\left( \frac{acd}{3} \right),z={{\log }_{4c}}\left( \frac{abd}{4} \right),\ w={{\log }_{5d}}\left( \frac{abc}{5} \right)\end{array} $$ and $$\begin{array}{l}\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+\frac{1}{w+1}={{\log }_{abcd}}N+1,\end{array} $$ the value of N is

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

Given:

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Solution:

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Option: (c)

Since $x = \log_{2a}\left(\frac{bcd}{2}\right)$

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

(\therefore\ \log_{abcd}2a = \frac{1}{x+1})

Similarly, $$\begin{array}{l}\frac{1}{y+1}={{\log }{abcd}}~3b,\frac{1}{z+1}={{\log }{abcd}}~4c\end{array} $$

\( \frac{1}{w+1} = \log_{abcd} 5d \)

(\therefore \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} + \frac{1}{w+1} = \log_{abcd}\left(2a \cdot 3b \cdot 4c \cdot 5d\right))

(\begin{array}{l}={{\log }_{abcd}}\left( 120abcd \right)\=4\end{array})

(\log_{abcd} (120+1))

(\begin{array}{l}={{\log }_{abcd}}(N+1)\end{array})

We have N = 120 when comparing

Problem 6: If $$a = \log_{12}{18}, b = \log_{24}{54}$$ then the value of $$ab + 5(a-b)$$ is

(a) 0
(b) 4
(c) 1
(d) None of these

Given:

Welcome to my site

Solution:

Welcome to my site

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

$b = \frac{1 + 3 \log_2 3}{3 + \log_2 3}$

Putting $$x = \log_2 3$$, we have

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

(\frac{6{{x}^{2}}+5x+1+5\left( -{{x}^{2}}+1 \right)}{\left( x+2 \right)\left( x+3 \right)} = \frac{{{x}^{2}}+5x+6}{\left( x+2 \right)\left( x+3 \right)} = 1)

Problem 7: (\begin{array}{l}\text{The value of }\ \frac{{{\log }_{96}}2 - {\log }_{2}24}{{{\log }_{12}}2 - {\log }_{2}192}\ \text{is}\end{array} )

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

Given:

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Solution:

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Option: (a)

Set log₂ 12 = a,

(\frac{1}{{{\log }_{2}}96}={{\log }_{96}}2={{\log }_{96}}({{2}^{3}}\times 12)=a+3)

(\begin{array}{l}{{\log }_{2}}24=1+a\{{\log }_{2}}192={{\log }_{2}}\left( 16\times 12 \right)=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}=\left( 1+a \right)\left( 3+a \right)-\left( 4+a \right)a\=3\end{array} )

Problem 8: The solution of the equation (\begin{array}{l}{{4}^{{{\log }_{2}}\log x}}=\log x-{{\left( \log x \right)}^{2}}+1\end{array} ) is: $\log x = \frac{1 \pm \sqrt{1-16 \log ^2 4}}{2}$.

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

Given:

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Solution:

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Option: (c)

log_2 log_x is meaningful if x > 1.

Since $$\begin{array}{l}{4^{\log_2 \log x}}= {2^{2\log_2 \log x}} = \left( {2^{\log_2 \log x}} \right)^2 = \left( \log x \right)^2 \end{array}$$

[\left[ a^{\log_a x} = x, a > 0, a \neq 1 \right]]

So the given equation reduces to:

(\begin{array}{l}2{{\left( \log x \right)}^{2}}-\log x-1=0 \ \log x = \frac{1 \pm \sqrt{1+8}}{4} \end{array} )

(\begin{array}{l} \Rightarrow x = 10, \quad x = \frac{1}{\sqrt{2}} \end{array} )

But for x > 1,

Log x > 0, therefore log x = 1, which implies 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, \infty))

(c) (-∞, 0)

(d) (0, 3)

Given:

This is a heading

Solution:

This is a heading

log_{0.5}(x - 1) < log_{0.25}(x - 1)

(\begin{array}{l}{{\log }_{0.5}}\left( x-1 \right)>{{\log }_{{{\left( 0.5 \right)}^{2}}}}\left( x-1 \right)\end{array})

(\begin{array}{l}\Rightarrow \frac{{{\log }_{0.5}}\left( x-1 \right)}{2}<{{\log }_{0.5}}{{\left( 0.5 \right)}^{2}}={\log }_{0.5}{\left( 0.25 \right)}\end{array})

$$\Leftrightarrow ,,,,x < 1 - {{0.5}^{\log_0.5(x-1)}}$$

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

Answer: (\frac{1}{{{\log }_{2}}2002}+\frac{1}{{{\log }_{3}}2002}+\frac{1}{{{\log }_{4}}2002}+…+\frac{1}{{{\log }_{2002}}2002})

Given:

This is bold text

Solution:

This is bold text

(\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+\…+\frac{1}{{{\log }_{2002}}n})

(\begin{array}{l}\frac{1}{{{\log }_{n}}2}+\frac{1}{{{\log }_{n}}3}+\frac{1}{{{\log }_{n}}4}+…+\frac{1}{{{\log }_{n}}2002} \end{array} ) , Since, (\begin{array}{l}{{\log }_{b}}a=\frac{1}{{{\log }_{a}}b}\end{array} )

$\log_{n}(2.3.4...2002)$

(\begin{array}{l}={{\log }_{n}}\left( 2002! \right) \ = \sum\limits_{k=1}^{2002} {\log_n k} \ = {{\log }_{n}}n + \sum\limits_{k=2}^{2002} {\log_n k} \ ={{\log }_{n}}n\end{array} )

Problem 11: If (x, y, z > 0) and such that (\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}), prove that (x^2y^2z^2=1).

Given:

This is a statement

Solution:

This is a statement

Let $$\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}=\lambda$$

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

‘\(x \log{x} + y \log{y} + z \log{z}\)’

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

$$\log {{x}^{x}}+\log {{y}^{y}}+\log {{z}^{z}}=0$$

(\Rightarrow \log \left( {{x}^{x}}{{y}^{y}}{{z}^{z}} \right)=\log \left( 1 \right)=0)

(\Rightarrow {{x}^{x}}{{y}^{y}}{{z}^{z}}=1 )

Problem 12: Solve: $$\log_3\left{5 + 4\log_3\left(x - 1\right)\right} = 2$$

Given:

This is an example

Solution:

This is an example

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

$$x > 1 \quad \text{and} \quad \log_3(x-1) > -\frac{5}{4}$$

(\begin{array}{l}x > 1 ,&&, x - 1 > \frac{3^{-5}}{4}\end{array})

(\begin{array}{l}\Rightarrow x > 1 , and , x > \frac{4}{3} \end{array})

(\begin{array}{l}\Rightarrow x > 1 + \frac{1}{3^{\frac{5}{4}}} \end{array}) he said

He said, “Now.”

log3\left({5 + 4\, log3\left(x - 1\right)}\right) = 2

(\begin{array}{l}5+4\log_3\left(x-1\right)=3^2\end{array})

$$\log_3(x-1) = 1$$

x = 4

x = 4

Clearly, x = 4 satisfies (i).

Therefore, x = 4 is the answer to the equation.

Solution 13: x = 9

Solution: Clearly, the given equation is meaningful, if $$\begin{array}{l}{3^{x}}-8 > 0 \Rightarrow 3^{x} > 8 \Rightarrow x > \log_3 8 \dots \dots (i)\end{array}$$ he said

He said, “Now.”

log_3 (3x - 8) = 2 - x

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

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

(\begin{array}{l}\Rightarrow {{3}^{2x}}-9\cdot {{3}^{x}}-1=0\end{array} )

(\begin{array}{l}\Rightarrow {{3}^{x}}-9=0,,,,,,,,,\left[ Since, {{3}^{x}}>8,,Therefore, ,,{{3}^{x}}-9 \ne 0 \right]\end{array} )

‘(\begin{array}{l} \Rightarrow x=2 \end{array})’

x = 2

Clearly, 2 > log₃ 8

Therefore, x = 2 is the answer to the equation.

Solution 14: x = e²

Given:

This is a heading

Solution:

This is a heading

It is evident that the equation is only applicable when x is greater than 0. he said

He said, “Now.”

2log_{x}(x) = 10x^2

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

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

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

(\Rightarrow 2\log^2 x - 2\log x - 1 = 0).

(\begin{array}{l}y=\frac{2\pm \sqrt{12}}{2}\end{array})

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

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

(\begin{array}{l}\Rightarrow x=10^{\pm1 \pm \sqrt{3}}\end{array} )

Problem 15: Solve: $$\log_2 (9-2x) = 10 \log (3-x)$$

Given:

This is a Heading

Solution:

This is a Heading

If we observe that the two sides of the given equation are meaningful,

9 - 2x > 0 AND 3 - x > 0

(\begin{array}{l} \Rightarrow 0 \leq x < 3 \end{array})

$x < \log_2 9 \ \text{and} \ x < 3$

x < 3 \quad\quad\quad\quad (i) she said

She said, “Now,”

log_2(9 - 2x) = 10 \cdot log(3 - x)

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

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

‘(\begin{array}{l}{2}^{x}=2\end{array})’

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

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

(\begin{array}{l} \Rightarrow y = 8 \ \text{or} \ y = 1 \end{array})

y = 9, 1

‘(\begin{array}{l}\Rightarrow \log_{2}{8}=x\end{array})’

x = 3, 0

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

Therefore, x = 0.

Problem 16: Solve: $$\left( x+1 \right)^{\log \left( x+1 \right)}=100\left( x+1 \right)$$

Given:

This is a statement

Solution:

This is a statement

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

He said, “Now.”

$$\left(x+1\right)^{\log\left(x+1\right)} = 100\left(x+1\right)$$

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

(\begin{array}{l}\log \left( x+1 \right)^2=\log 100+\log \left( x+1 \right)\end{array})

(\left{ \log {{\left( x+1 \right)}^{2}} \right} = 2 + \log \left( x+1 \right) \Rightarrow \log {{\left( x+1 \right)}^{2}} = 2 + \log \left( x+1 \right))

$$y^2 - y - 2 = 0, \text{where } y = \log(x+1)$$

y = 2, -1

$$\Rightarrow \log_{10}(x+1) = 2 \text{ and } \log_{10}(x+1) = -1$$

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

x = 99, x = -0.9

Problem 17: Evaluate $\sqrt[3]{72.3},,,if,,\log 0.723=\overline{1}.8591$.

Given text:

This is a heading

Solution:

This is a heading

Then, (x = 4.3).

log x = log(72.3)^(1/3)

‘(\begin{array}{l}\Rightarrow \log x = \log 72.3^{\frac{1}{3}}\end{array} )’

(\Rightarrow \log x = 0.6030)

‘(\begin{array}{l} \Rightarrow x = 10^{0.6197} \end{array} )’

(\begin{array}{l}\Rightarrow x=\log_{10} \left( 0.6197 \right)\end{array} )

x = 4.166

Answer:

Problem 18: (\sqrt[5]{10076} = 4.01 )

Given:

Hello World

Solution:

Hello World

Then, $x = 4.08$.

log x = log (10076)^1/5

‘(\begin{array}{l}\log x = \log 10076^{\frac{1}{5}}\end{array} )’

(\begin{array}{l}\Rightarrow \log x = 0.80058\end{array})

‘(\log x = 0.8058)’

(\begin{array}{l}\Rightarrow x=\log_{10} \left( 0.8058 \right)\end{array} )

x = 6.409

Problem 19: What is the logarithm of $32\sqrt[5]{4},,to,the,base,,2\sqrt{2}$?

Given:

This is a heading

Solution:

This is a heading

Here we can write $$32\sqrt[5]{4},,as,,{{2}^{5}}{{4}^{1/5}}={{\left( 2 \right)}^{27/5}},and,2\sqrt{2},,as,{{2}^{\frac{3}{2}}}$$

We can solve it by using the formula (\begin{array}{l}{{\log }_{a}}{{M}^{x}}=x{{\log }_{a}}.M,and,{{\log }_{{{a}^{x}}}}M=\frac{1}{x}{{\log }_{a}}M\end{array} ).

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

Problem 20: Prove that, $$\log_{\frac{4}{3}}\left( 1.\overline{3} \right)=1$$

Given:

This is a heading

Solution:

This is a heading

We get (\begin{array}{l}1.\overline{3}=\frac{4}{3},\end{array} ) by solving, and we can use the formula (\begin{array}{l}{{\log }_{a}}a=1.\end{array} )

(\begin{array}{l}{\log_{\frac{4}{3}}1.\overline{3}}=1\end{array})

Let $x = 1.333 \dots (i)$

10x = 13.3333….

10x = 13.3333….

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

So $\begin{array}{l}9x=12\Rightarrow x=\frac{12}{9},x=\frac{4}{3}\end{array}$

Now $$\log_{\frac{4}{3}} \frac{1}{\overline{3}}=\log_{\frac{4}{3}} \left( \frac{4}{3} \right)=1$$

Problem 21: If

(\underset{N\to \infty }{\mathop{\lim }},\left[ {{\left( {{\log }_{2}}N \right)}^{-1}}+{{\left( {{\log }_{3}}N \right)}^{-1}}+…+{{\left( {{\log }_{n}}N \right)}^{-1}} \right]) is the limit as (N) approaches infinity of (\left[ {{\left( {{\log }_{2}}N \right)}^{-1}}+{{\left( {{\log }_{3}}N \right)}^{-1}}+…+{{\left( {{\log }_{n}}N \right)}^{-1}} \right]) when (N=n!), (n \in N), and (n\ge 2).

Given:

This is a statement

Solution:

This is a statement

We can write the given expansion as $$\log_{a}b = \frac{1}{\log_{b}a}$$

(\begin{array}{l}{{\log }_{n!}}2+{{\log }_{n!}}3+……+{{\log }_{n!}}n\end{array} ) and then by using (\begin{array}{l}{{\log }_{a}}\left( n!^2 \right)={{\log }_{a}}n!+{{\log }_{a}}n!\end{array} )

(\begin{array}{l}\Rightarrow {{\left( {{\log }_{2}}N \right)}^{-1}}+{{\left( {{\log }_{3}}N \right)}^{-1}}+\dots+{{\left( {{\log }_{n}}N \right)}^{-1}}\={{\log }_{N}}2+{{\log }_{N}}3+\dots+{{\log }_{N}}n={{\log }_{n}}\left( 2\cdot 3 \cdot \dots \cdot N \right)={{\log }_{N}}N=1.\end{array})

Problem 22: (\begin{array}{l}\text{If}\ \log {{x}^{2}}-\log 2x=3\log 3-\log 6\ \text{ then x equals} \\ \text{x = }\sqrt{6}\end{array})

Given:

This is a test

Solution:

This is a test

By using $$\log_a(M.N) = \log_a M + \log_a N \text{ and } \log_a M^x = x \log_a M$$

Clearly $x > 0$. Then the given equation can be written as $x^2 + 2x + 1 = 0$.

(\begin{array}{l}\log\left(x\right)=2\log\left(3\right)\Rightarrow x=9\end{array})

Solution to Problem 23:

Let (x = \log_{2-\sqrt{3}} (2+\sqrt{3})). Then,

(2+\sqrt{3} = (2-\sqrt{3})^x \Rightarrow (2+\sqrt{3})^2 = (2-\sqrt{3})^{2x} \Rightarrow 4+6 = 4-6x \Rightarrow 6x = -2 \Rightarrow x = -1)

Therefore, (\log_{2-\sqrt{3}} (2+\sqrt{3}) = -1).

Given:

Hi

Solution:

Hi

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

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

We can easily prove this by using $\log_{1/N} N = -1$.

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

Example 24: Prove that, $$\log _{5}\sqrt{5\sqrt{5\sqrt{5 \dots \infty}}} = 1$$

Given:

This is an example

Solution:

This is an example

Here (\begin{array}{l}y=\sqrt{5y}\end{array} ), where ( \begin{array}{l}y=\sqrt{5\sqrt{5\sqrt{5……..\infty }}}\end{array} ) can be represented as (\sqrt{5\sqrt{5\sqrt{5……..\infty }}}).

Therefore, by finding the value of y we can confirm this.

‘Let (y = \sqrt{5^{\sqrt{5^{\sqrt{5^{\ddots}}}}})’

(\begin{array}{l}\Rightarrow {{y}^{2}}-5y=0 \\ \Rightarrow y(y-5)=0 \\ \Rightarrow y=0,,,,,,,,,,,,,,,,,,,,,,,,,or,,,,,,,,,,,,,,,,,,,,,,,,,y=5 \end{array})

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

\(\therefore \log_5 5 = 1\)

Problem 25: Prove that, $$\log_{2.25}(0.\overline{4})=-1$$

Given: This is a sentence.

Solution: This is a sentence.

(\begin{array}{l}{{\log }_{1/N}}N=-1\end{array} \Rightarrow N=\frac{1}{e^{-1}} \Rightarrow N = \frac{1}{\frac{1}{e}} \Rightarrow N = e)

x = 0.4444 ... (i)

10x = 4.4444…. (ii)

10x = 4.4444…. (ii)

Equation (ii) - Equation (i)

So $$9x=4 \Rightarrow x=\frac{4}{9}$$

Also, $$2.25=\frac{225}{100}=\frac{9}{4};,,,,,,,,,,,,,,{{\log }{2.25}}\left( 0.\overline{4} \right)={{\log }{\left( \frac{9}{4} \right)}}\left( \frac{4}{9} \right)=-1$$

Problem26: Find the value of $$2^{\log_{6}18} \cdot 0.3^{\log_{6}3}$$

Given:

This is a header

Solution:

This is a header

We can solve the above problem step by step by using the following equations:

$$\begin{array}{l}{{\log }_{a}}\left( M.N \right)={{\log }_{a}}M+{{\log }_{a}}N,and,,{{a}^{{{\log }_{e}}c}}={{c}^{{{\log }_{e}}a}}\end{array}$$

(\begin{array}{l} \Rightarrow {{2}^{{{\log }_{6}}18}}{{\left( 3 \right)}^{{{\log }_{6}}3}}={{2}^{{{\log }_{6}}\left( 6\times 3 \right)}}{{.3}^{{{\log }_{6}}3}}\ ={{2}^{1+{{\log }_{6}}3}}{{.3}^{{{\log }_{6}}3}}={{2.2}^{{{\log }_{6}}3}}{{.3}^{{{\log }_{6}}3}},,,\left( Since, {{a}^{{{\log }_{6}}c}}\={{c}^{{{\log }_{6}}a}} \right) \end{array})

= \( \begin{array}{l} 2.\left( 3 \right)^{\log_6 2}.\left( 3 \right)^{\log_6 3} = 2\left( 3 \right)^{\log_6 2 + \log_6 3} = 2\left( 3 \right)^{\log_6 6} = 2\left( 3 \right) = 6 \end{array} \)

Problem 27: Find the value of, $$\log_{\sec ,,\alpha}\left( {\cos }^{3}\alpha \right) \text{where} \alpha \in \left( 0,\frac{\pi}{2} \right)$$

Solution: Consider $$\log_{\sec ,,\alpha}\left(\cos^3\alpha\right)=x.$$

Therefore, by using the formula (y={{\log }_{a}}x\Leftrightarrow {{a}^{y}}=x), we can write (\begin{array}{l}co{{s}^{3}}\alpha ={{\left( \sec ,,\alpha \right)}^{x}}.\end{array} ) Hence, by solving this equation, we can get the value of $x$.

Let $$\log_{\sec ,,\alpha} \cos^3\alpha = x$$

$$\cos^3\alpha = \left(\sec\alpha\right)^x \Rightarrow \left(\cos\alpha\right)^3 = \left(\frac{1}{\cos\alpha}\right)^x \Rightarrow \left(\cos\alpha\right)^3 = \left(\cos\alpha\right)^{-x} \Rightarrow x = -3$$

Problem 28: If $k \in \mathbb{N}$ such that $\log_2 x + \log_4 x + \log_8 x = \log_k x$ and $\forall x \in \mathbb{R’}$, if $k = a^{1/b}$ where $a \in \mathbb{N}$ and $b \in \mathbb{N}$ is a prime number, then find the value of $a + b$.

Given:

This is a heading

Solution:

This is a heading

By using $$\log_b a = \frac{\log_c a}{\log_c b} = \frac{\log a}{\log b}$$

By comparing (k={{\left( a \right)}^{1/b}}) to the value of k, we can obtain the values of both a and 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}\left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3} \right]=\frac{\log x}{\log 2}\left( \frac{11}{6} \right)=\frac{\log x}{\log k} \ \Rightarrow \log x\left[ \frac{11}{6}\frac{1}{\log 2}-\frac{1}{\log k} \right]=0 \end{array}

\Rightarrow \log x\left[ \frac{11}{6\log 2}-\frac{1}{\log k} \right]=0

Also, $$\begin{array}{l}\frac{11}{6}\frac{1}{\log 2}-\frac{1}{\log k}=0\Rightarrow \frac{11}{6}=\frac{\log 2}{\log k}\Rightarrow \frac{11}{6}={{\log }_k}2\end{array} $$

So $$2={{k}^{\frac{11}{6}}};{{2}^{6/11}}=k\Rightarrow {{\left( {{2}^{6}} \right)}^{\frac{1}{11}}}=k\Rightarrow {{\left( 64 \right)}^{\frac{1}{11}}}=k$$

(\begin{array}{l}\text{Comparing using }k={{\left( a \right)}^{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}}],=,2\end{array})

Given:

This is a heading

Solution:

This is a heading

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

Problem 30: In the expansion of (\begin{array}{l}2{{\log}_{e}}x-{{\log}_{e}}(x+1)-{{\log}_{e}}(x-1)\end{array} ), what is the coefficient of x-4?

Given:

This is a Header

Solution:

This is a Header

(\begin{array}{l}2{{\log}_{e}}x-{{\log}_{e}}\left{\left( 1+\frac{1}{x} \right)x \right}-{{\log }_{e}}\left{\left( 1-\frac{1}{x} \right)x \right}\=2{{\log}_{e}}x-\left{{{\log}_{e}}\left(1+\frac{1}{x} \right)+{{\log}_{e}}x \right}-\left{ {{\log}_{e}}\left(1-\frac{1}{x} \right)+{{\log}_{e}}x \right}\=-\left{{{\log}_{e}}\left(1+\frac{1}{x} \right)+{{\log }_{e}}\left(1-\frac{1}{x} \right) \right}\=2\left{\frac{1}{2{{x}^{2}}}+\frac{1}{4{{x}^{4}}}+……. \right}\ \text{The coefficient of}\ {{x}^{-4}}=2.\frac{1}{4}=\frac{1}{2}\\end{array})

Problem 31: (\displaystyle \sum_{i=2}^{n-1} \log_e \left(1+\frac{1}{i}\right) = )

Given:

This is a statement

Solution:

This is a statement

(\begin{array}{l}{{\log}_{e}}2+{{\log}_{e}}\left(\frac{3}{2} \right)+{{\log}_{e}}\left( \frac{4}{3} \right)+….+{{\log}_{e}}\left(\frac{n}{n-1} \right)\ ={{\log }_{e}}2+{{\log}_{e}}3-{{\log}_{e}}2+{{\log}_{e}}4-{{\log}_{e}}3+…… +{{\log}_{e}}(n)-{{\log}_{e}}(n-1)\ ={{\log}_{e}}n.\\end{array})

Problem 32: The coefficient of ${{x}^{n}}$ in the expansion of $\log_{e}(1+3x+2{{x}^{2}})$ is

Given:

This is a heading

Solution:

This is a heading

(\begin{array}{l} \log (1+3x+2{{x}^{2}})=\log (1+x)+\log (1+2x)\ =\sum\limits_{n=1}^{\infty}{{{(-1)}^{n-1}}}\frac{{{x}^{n}}}{n}+\sum\limits_{n=1}^{\infty}{{{(-1)}^{n-1}}}\frac{{{(2x)}^{n}}}{n}\ = \sum\limits_{n=1}^{\infty}{{{(-1)}^{n-1}}}\left(\frac{1}{n}+\frac{{{2}^{n}}}{n} \right),,{{x}^{n}} \ =\sum\limits_{n=1}^{\infty}{{{(-1)}^{n-1}}}\left(\frac{1+{{2}^{n}}}{n} \right),,{{x}^{n}}\ \text{Therefore, the coefficient of}\ {{x}^{n}}={{(-1)}^{n-1}}\left(\frac{{{2}^{n}}+1}{n} \right)\ \end{array})

‘\(\begin{array}{l}[,,{{(-1)}^{n}}={{(-1)}^{n+2}}={{(-1)}^{n+4}}=\dots]\\end{array} \)’

Problem 33: (\displaystyle \sum_{n=1}^{\infty}\left(\frac{1}{2n}+\frac{1}{2n+1}\right)\frac{1}{4^n} = )

Given:

This is an example

Solution:

This is an example

(\begin{array}{l}S=\left{1+\frac{{{\left(\frac{1}{2} \right)}^{2}}}{2}+\frac{{{\left(\frac{1}{2} \right)}^{4}}}{4}+ \dots \right} + 2\left{\frac{1}{2}+\frac{{{\left( \frac{1}{2} \right)}^{3}}}{3}+\frac{{{\left( \frac{1}{2} \right)}^{5}}}{5\ }+ \dots \right} - 1\ =1-\frac{1}{2}{{\log }_{e}}\left(1+\frac{1}{2} \right)\text{ }\left(1-\frac{1}{2} \right)+{{\log }_{e}}\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}} \right)-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) \text {is equal to}\ (1) x^{2}+\frac{1}{2} x^{4}+\frac{1}{3} x^{6}+\ldots \ldots \infty\ (2) \frac{1}{x^{2}}+\frac{1}{2 x^{4}}+\frac{1}{3 x^{6}}+\ldots \ldots \infty\ (3) -\left(\frac{1}{x^{2}}+\frac{1}{2 x^{4}}+\frac{1}{3 x^{6}}+\ldots \ldots \infty\right)\ (4) \text {None of these}\ \text{Solution}:\ \begin{array}{l} 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 \left(x^{2}-1\right)=\log \frac{x^{2}}{x^{2}-1} \ =-\log \left(\frac{x^{2}-1}{x^{2}}\right) \quad=-\log \left(1-\frac{1}{x^{2}}\right) \ =-\left[\frac{-1}{x^{2}}-\frac{1}{2}\left(\frac{1}{x^{2}}\right)^{2}-\frac{1}{3}\left(\frac{1}{x^{2}}\right)^{3}-\frac{1}{4}\left(\frac{1}{x^{2}}\right)^{4} \ldots \ldots \ldots \infty\right. \ =\frac{1}{x^{2}}+\frac{1}{2 x^{4}}+\frac{1}{3 x^{6}}+\frac{1}{4 x^{8}}+\ldots \ldots \ldots \infty \end{array}\ Answer: [2]\end{array} )

Solution:

2 $\log x - \log(x+1) - \log(x-1)$ is equal to

(1) $x^2 + \frac{1}{2}x^4 + \frac{1}{3}x^6 + \ldots \ldots \infty$

(2) $\frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \ldots \ldots \infty$

(3) $-\left(\frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \ldots \ldots \infty\right)$

(4) None of these

Solution:

$\begin{array}{l} 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 \left(x^{2}-1\right)=\log \frac{x^{2}}{x^{2}-1} \ =-\log \left(\frac{x^{2}-1}{x^{2}}\right) \quad=-\log \left(1-\frac{1}{x^{2}}\right) \ =-\left[\frac{-1}{x^{2}}-\frac{1}{2}\left(\frac{1}{x^{2}}\right)^{2}-\frac{1}{3}\left(\frac{1}{x^{2}}\right)^{3}-\frac{1}{4}\left(\frac{1}{x^{2}}\right)^{4} \ldots \ldots \ldots \infty\right. \ =\frac{1}{x^{2}}+\frac{1}{2 x^{4}}+\frac{1}{3 x^{6}}+\frac{1}{4

Problem 35: \begin{array}{l} \text{The coefficient of} \ n^{-r}\ \text{in the expansion of }\ \log_{10}\left(\frac{n}{n-1}\right)\ (1) \frac{1}{r \log_{e} 10}\ (2) \frac{\log_{e} 10}{r}\ (3) -\frac{\log_{e} 10}{r}\ (4) \text { None of these}\ \text{Solution}: \ \log_{10}\left(\frac{n}{n-1}\right) \Rightarrow \log_{e}\left(\frac{n}{n-1}\right) \cdot \log_{10} e\ \begin{array}{l} \Rightarrow-\log \left(\frac{n-1}{n}\right) \log_{10} e \ \Rightarrow-\log \left(1-\frac{1}{n}\right) \log_{10} e \ =\left[\frac{1}{n}+\frac{1}{2 n^{2}}+\frac{1}{3 n^{3}}+\ldots \ldots+\frac{1}{m^{r}}+\ldots \ldots \ldots \infty\right] \log_{10} e \end{array}\ \therefore \quad \text {The coefficient of }\ n^{-r}=\frac{1}{r} \log_{10} e\ =\frac{1}{r \log_{e} 10}\ Answer: [1] \end{array}

Problem 36: $$\log \frac{(1+x)^{(1-x) / 2}}{(1-x)^{(1+x) / 2}} \text{is equal to}$$

(1) $x+\frac{5 x^{3}}{2.3}+\frac{9 x^{5}}{4.5}+\frac{13 x^{7}}{6.7}+\ldots \ldots . .+\infty$

(2) $x-\frac{5 x^{3}}{2.3}+\frac{9 x^{5}}{4.5}-\frac{13 x^{7}}{6.7}+\ldots \ldots . .+\infty$

(3) $x-\frac{5 x^{3}}{2.3}-\frac{9 x^{5}}{4.5}-\frac{13 x^{7}}{6.7}-\ldots \ldots . .-\infty$

(4) $\text{None of these}$

Solution: $$\log \frac{(1+x)^{(1-x) / 2}}{(1-x)^{(1+x) / 2}}$$ $$=\frac{1}{2}(1-x) \log (1+x)-\frac{1}{2}(1+x) \log (1-x)$$ $$=\frac{1}{2}[\log (1+x)-\log (1-x)]-\frac{1}{2}[\log (1+x)+\log (1-x)]$$ $$=\frac{1}{2} \cdot 2\left[\left[x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots \ldots .\right]-\frac{1}{2} \cdot x(-2)\left[\frac{1}{2} x^{2}+\frac{x^{4}}{4}+\ldots . .\right]\right.$$ $$=x+\left(\frac{1}{3}+\frac{1}{2}\right) x^{2}+\left(\frac{1}{5}+\frac{1}{4}\right) x^{5}+\left(\frac{1}{7}+\frac{1}{6}\right) x^{7}+\ldots \ldots . $$ $$=x+\frac{5 x^{3}}{2.3}+\frac{9 x^{5}}{4.5}+\frac{13 x^{7}}{6.7}+\ldots \ldots \ldots . $$

Answer: [1]

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Frequently Asked Questions

Who invented logarithms?

Logarithms were invented by John Napier in 1614.

John Napier invented logarithms.

The Product Rule of Logarithms states that: $$log_a(xy) = log_a(x) + log_a(y)$$

log_a(m\cdot n) = log_a(m) + log_a(n)

The quotient rule of logarithms states that: $$\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)$$

log_a(m/n) = log_am - log_an

1. Logarithms are used to calculate the pH of a solution.
2. Logarithms are used to calculate the decibel level of a sound.

Logarithms are utilized by biologists to calculate population growth rates.

It is also used to measure the magnitude of earthquakes.

The Power Rule of Logarithms states that $\log_b x^n = n \log_b x$

log(x^a) = a * log(x)