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: ${\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a}$

($$\begin{array}{l}{a}^{{\log }_{a}b}=b,a>0,a\ne 1\text{and}b>0\end{array}$$)

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 $a^{-\mathrm{\infty }}=0$, i.e. ${\log }_{a}0=-\mathrm{\infty }$ (if $a>1$).

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

Corollary 5:

If $a>1$, then ${\log }_{a}b\to \mathrm{\infty }$

If 0 < a < 1, then ${\log }_{a}b\to -\mathrm{\infty }$ and ${\log }_{a}b\to \mathrm{\infty }$

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. $${\log }_{a}1=0\text{(for }a>0,a\ne 1)$$

5. $${\log }_{a}a=1\text{(where }a>0\text{ and }a\ne 1\text{)}$$

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

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

($$\begin{array}{l}0<a<1,{\log }_{a}x=\{\begin{array}{l}+ve,0<x<10,x=1-ve,x>1\end{array}\end{array}$$ )

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:

This is a Header

base tan 450 = 1

log is not defined.

Example 2: ($$\begin{array}{l}\text{Find the value of }lo{g}_{(se{c}^2{60}^0-ta{n}^2{60}^0)}cos{60}^0\end{array}$$ )

Given:

This is a heading

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 $${\log }_{({\sin }^2{30}^{\circ }+{\cos }^2{30}^{\circ })}1={\log }_{1}1\ne 1$$

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

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

${\log }_a\left(\prod _{i=1}^n{x}_i\right)=\sum _{i=1}^n{\log }_a{x}_i,\mathrm{\forall }{x}_i>0$

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. $${\log }_{a^{\beta }}m=\frac{1}{\beta }{\log }_{a}m$$

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

6. $${\log }_{b}a{\log }_{a}b=1\iff {\log }_{b}a=\frac{1}{{\log }_{a}b}$$

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

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

3. $${\log }_{a^k}x=\frac{1}{k}{\log }_{a}x,a>0,a\ne 1,x>0$$

4. $${\log }_a{x}^{2k}=2k{\log }_a\left|x\right|,a>0,a\ne 1,k\in \mathbb{I}$$

5. $${\log }_{a^{2k}}x=\frac{1}{2k}{\log }_{|a|}x,x>0,a\ne \pm 1,k\in \mathbb{I}10$$

6. ${\log }_{a^{\alpha }}{x}^{\beta }=\frac{\beta }{\alpha }{\log }_{a}x$, $x>0$, $a>0$, $a\ne 1$, $\alpha \ne 0$

7. ${\log }_a{x}^2\ne 2{\log }_{a}x,a>0,a\ne 1$

The domain of loga (x)^2 is $\mathbb{R}\setminus 0$, whereas the domain of loga x is $(0,\mathrm{\infty })$, 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:

This is a test

($$\begin{array}{l}\Rightarrow {\log }_5;(3.x^2+2^x)={\log }_5;64\end{array}$$ )

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

By extra property (i)

($$\begin{array}{l}\Rightarrow {\log }_{5}x=\frac{{\log }_{4.2}64}{{\log }_{4.2}4.2}\end{array}$$ )

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

$$\therefore {\log }_{5}x=4$$

(\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:

This is a headline

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

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

Now, let $\lambda =12$, then

($$\begin{array}{l}LHS={\log }_{2}2\lambda +{\log }_{2}8\lambda -{\log }_{2}16\lambda -{\log }_2\lambda \end{array}$$ )

($$\begin{array}{l}=({\log }_{2}2\cdot {\log }_{2}8)-\left({\log }_{2}16\right){\log }_2\lambda +({\log }_{2}2+{\log }_{2}8+{\log }_{2}16){\log }_2\lambda \end{array}$$ )

($$\begin{array}{l}=({\log }_{2}2+{\log }_2\lambda )({\log }_{2}8+{\log }_2\lambda )-({\log }_{2}16+{\log }_2\lambda ){\log }_2\lambda \end{array}$$ )

($$\begin{array}{l}=(1+{\log }_2\lambda )(3{\log }_{2}2+{\log }_2\lambda )\text{-}(4{\log }_{2}2+{\log }_2\lambda ){\log }_2\lambda \end{array}$$ )

($$\begin{array}{l}=(1+{\log }_2\lambda )(3+{\log }_2\lambda )-4{\log }_2\lambda -{{\log }_2}^2{\lambda }^2\end{array}$$)

3 =

RHS = Right Hand Side

Properties of Monotonicity of Logarithm

Logarithm with a Constant Base

1. $${\log }_{a}x>{\log }_{a}y\iff \{\begin{array}{lll}x>y>0,& \text{if }a>10<x<y,& \text{if }0<a<1\end{array}$$

2. $${\log }_{a}x<{\log }_{a}y⟺\{\begin{array}{lll}0<x<y,& \text{if }a>1x>y>0,& \text{if }0<a<1\end{array}$$

3. ${\log }_{a}x>p\iff \{\begin{array}{lll}x>a^p,& \text{if }a>10<x<a^p,& \text{if }0<a<1\end{array}$

$4.{\log }_{a}x<p\iff \{\begin{array}{lll}0<x<a^p,& \text{if }a>1x>a^p,& \text{if }0<a<1\end{array}$

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 $(0,1)\cup (1,\mathrm{\infty })$

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 $1\le x<2$, $[x]=1⟹{\log }_a[x]=0$

When $$2\le x<3,x=2\Rightarrow {\log }_{a}x={\log }_{a}2$$

When $3\le x<4$, $x=3\Rightarrow {\log }_{a}x={\log }_{a}3$ 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. $$|{\log }_{b}a+{\log }_{a}b|\ge 2,\mathrm{\forall }a>0,a\ne 1,b>0,b\ne 1$$

5. ($$\begin{array}{l}{\log }_2{\log }_2\underset{n,,times}{\underbrace{\sqrt[2^n]{2}}}=-n\end{array}$$ )

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 $$x={\log }_{c}b+{\log }_{b}c,y={\log }_{a}c+{\log }_{c}a,z={\log }_{a}b+{\log }_{b}a,$$ 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:

This is a heading

Solution:

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x > 2

⇒ x > 2 \ (i)

And $$\begin{array}{l}{\log }_2(x-2)<{\log }_{2^2}(x-2)=\frac{1}{2}{\log }_2(x-2)\end{array}$$

($$\begin{array}{l}\Rightarrow {\log }_2(x-2)<\frac{1}{2}{\log }_2(x-2)\end{array}$$ )

($$\begin{array}{l}\Rightarrow \frac{1}{2}{\log }_2(x-2)<0\\ \Downarrow 2{\log }_2(x-2)<0\\ \Downarrow {\log }_2(x-2)<0\end{array}$$ )

‘($$\begin{array}{l}x-2<2\end{array}$$)’

x - 2 < 1

⇒ x < 3 \ \ (ii)

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

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

Problem 3: If $$\begin{array}{l}{a}^{{\log }_{b}c}={3.3}^{{\log }_{4}3}{{.3}^{{\log }_{4}3}}^{{}^{{\log }_{4}3}}{.3}^{{\log }_4{3}^{{}^{{\log }_4{3}^{{\log }_{b}c}}}}\dots \mathrm{\infty }\end{array}$$

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:

This is a header

Option: (c)

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

($$\begin{array}{l}=3^{{\log }_{4/3}4}=3^{{\log }_4\left(4/3\right)}=3^{1/(1-{\log }_{4}3)}\end{array}$$)

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)

($$\begin{array}{l}{\log }_3(x^2-4x+3)=3^{(x-3)}\dots \dots \dots .(i)\end{array}$$)

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 $$x\in \mathrm{\Phi }$$.

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))

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

(\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

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

(\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)

($$\text{Extra close brace or missing open brace}$$ )

And $\begin{array}{l}\frac{1}{{\log }_{12}2}={\log }_{2}12=a.\end{array}$

Therefore, the given expression

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

Problem 8: The solution of the equation ($$\begin{array}{l}{4}^{{\log }_2\log x}=\log x-{(\log x)}^2+1\end{array}$$ ) 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:

This is a headline

Option: (c)

log_2 log_x is meaningful if x > 1.

Since $$\begin{array}{l}{4}^{{\log }_2\log x}=2^{2{\log }_2\log x}={\left(2^{{\log }_2\log x}\right)}^2={(\log x)}^2\end{array}$$

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

So the given equation reduces to:

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

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

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)

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

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

$$\iff ,,,,x<1-{0.5}^{{\log }_0.5(x-1)}$$

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

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

Given:

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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})

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

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

($$\begin{array}{l}={\log }_n(2002!)=\sum _{k=1}^{2002}{\log }_{n}k={\log }_{n}n+\sum _{k=2}^{2002}{\log }_{n}k={\log }_{n}n\end{array}$$ )

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

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

‘$x\log x+y\log y+z\log z$’

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

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

(\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: $$\text{Missing or unrecognized delimiter for left}$$

Given:

This is an example

Solution:

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

$$x>1\text{and}{\log }_3(x-1)>-\frac{5}{4}$$

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

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

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

He said, “Now.”

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

($$\begin{array}{l}5+4{\log }_3(x-1)=3^2\end{array}$$)

$${\log }_3(x-1)=1$$

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

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

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

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

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

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

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

($$\text{Missing or unrecognized delimiter for left}$$)

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

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

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

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

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

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

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

Problem 15: Solve: $${\log }_2(9-2x)=10\log (3-x)$$

Given:

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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

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

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

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

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.”

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

($$\text{Missing or unrecognized delimiter for left}$$ )

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

(\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 $\sqrt[3]{72.3},,,if,,\log 0.723=\bar{1}.8591$.

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, $${\log }_{\frac{4}{3}}\left(1.\bar{3}\right)=1$$

Given:

This is a heading

Solution:

This is a heading

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

($$\begin{array}{l}{\log }_{\frac{4}{3}}1.\bar{3}=1\end{array}$$)

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 $${\log }_{\frac{4}{3}}\frac{1}{\bar{3}}={\log }_{\frac{4}{3}}\left(\frac{4}{3}\right)=1$$

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 $${\log }_{a}b=\frac{1}{{\log }_{b}a}$$

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

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

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 ${\log }_{1/N}N=-1$.

($$\begin{array}{l}\Rightarrow -1.{\log }_{2-\sqrt{3}}(2-\sqrt{3})=-1\end{array}$$)

Example 24: Prove that, $${\log }_5\sqrt{5\sqrt{5\sqrt{5\dots \mathrm{\infty }}}}=1$$

Given:

This is an example

Solution:

This is an example

Here ($$\begin{array}{l}y=\sqrt{5y}\end{array}$$ ), where ( $$\begin{array}{l}y=\sqrt{5\sqrt{5\sqrt{5\dots \dots ..\mathrm{\infty }}}}\end{array}$$ ) 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}}}}})’

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

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

\(\therefore \log_5 5 = 1\)

Problem 25: Prove that, $${\log }_{2.25}(0.\bar{4})=-1$$

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:

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Solution:

This is a header

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

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

($$\begin{array}{l}\Rightarrow 2^{{\log }_{6}18}{\left(3\right)}^{{\log }_{6}3}=2^{{\log }_6(6\times 3)}{.3}^{{\log }_{6}3}=2^{1+{\log }_{6}3}{.3}^{{\log }_{6}3}={2.2}^{{\log }_{6}3}{.3}^{{\log }_{6}3},,,(Since,a^{{\log }_{6}c}\text{=}{c}^{{\log }_{6}a})\end{array}$$)

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

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

Solution: Consider $${\log }_{\sec ,,\alpha }({\cos }^3\alpha )=x.$$

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

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

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

Problem 28: If $k\in \mathbb{N}$ such that ${\log }_{2}x+{\log }_{4}x+{\log }_{8}x={\log }_{k}x$ and $\mathrm{\forall }x\in {\mathbb{R}}^{\prime }$, 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, $$\begin{array}{l}\frac{\log x}{\log 2}+\frac{\log x}{2\log 2}+\frac{\log x}{3\log 2}=\frac{\log x}{\log k}\Rightarrow \frac{\log x}{\log 2}[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}]=\frac{\log x}{\log 2}\left(\frac{11}{6}\right)=\frac{\log x}{\log k}\Rightarrow \log x[\frac{11}{6}\frac{1}{\log 2}-\frac{1}{\log k}]=0\end{array}$$

\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:

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Solution:

This is a heading

$$\text{Missing or unrecognized delimiter for left}$$

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:

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(\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 ${\log }_e(1+3x+2x^2)$ is

Given:

This is a heading

Solution:

This is a heading

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

‘$\text{Missing end{array}}$’

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

($$\text{Missing or unrecognized delimiter for left}$$ )

Problem 34: ($$\text{Missing end{array}}$$\ Answer: [2]\end{array} )

Solution:

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

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

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

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

(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: $$\text{Missing end{array}}$$\ \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) $x+\frac{5x^3}{2.3}+\frac{9x^5}{4.5}+\frac{13x^7}{6.7}+\dots \dots ..+\mathrm{\infty }$

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

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

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

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

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: $$lo{g}_a(xy)=lo{g}_a(x)+lo{g}_a(y)$$

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)