Introduction to Logarithm

  

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 ${{a}^{x}}=b\left( where\,\,a>0,\ne 1 \right),$ then x is called the logarithm of b to the base a and we write loga b = x, clearly b > 0. Thus ${{\log }_{a}}b=x\Leftrightarrow {{a}^{x}}=b,a>0,a\ne 1$ and b > 0.

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

i.e. log10 e = 0.4343

∴ ln b = 2.303 log10 b

Since, $\left[\ln \,b={{\log }_{10}}b\times {{\log }_{e}}10=\frac{1}{{{\log }_{10}}e}\times {{\log }_{10}}b \right.\left. =\frac{1}{0.4343}{{\log }_{10}}b=2.303\ {{\log }_{10}}b \right]$

⇒ Important Points

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

Laws of Logarithm

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

${{a}^{{{\log }_{a}}b}}=b,a>0,a\ne 1\,and\,b>0$

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 called logarithmic function. In domain is $\left( 0,\infty \right)$ and range is R (set of all real numbers).

Corollary 3: ${{a}^{x}}>0,\forall x\in R$

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

Corollary 4:

1. If a > 1, then ${{a}^{-\infty }}=0$ i.e. $\,\,\,\,\,{{\log }_{a}}0=-\infty\left( if\,a>1 \right)$
2. If 0 < a < 1, then ${{a}^{\infty }}=0$ i.e. $\,\,\,\,\,{{\log }_{a}}0=+\infty \left( if\,0<a<1 \right)$

Corollary 5:

• ${{\log }_{a}}b\to \infty ,if\,a>1,b\to \infty$
• ${{\log }_{a}}b\to -\infty ,if\,0<a<1,b\to \infty$

Remarks

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

If $a>1,{{\log }_{a}}x=\left\{ \begin{matrix} +ve,\,\,\,\,\,\,\,\,\,\,\,\,\,x>1 \\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=1 \\ -ve,\,\,\,\,\,0<x<1 \\ \end{matrix} \right.$

And if,

$0<a<1,{{\log }_{a}}x=\left\{ \begin{matrix} +ve,\,\,\,\,\,0<x<1 \\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=1 \\ -ve,\,\,\,\,\,\,\,\,\,\,\,\,\,x>1 \\ \end{matrix} \right.$

Solved Examples on Logarithm

Example: 1: Find the value of ${{\log }_{\tan 45{}^\circ }}\cot 30{}^\circ$

Solution:

Here, base tan 450 = 1

∴ log is not defined.

Example: 2: Find the value of $log_{(sec^2 60^0 – tan^2 60^0)} cos 60^0$

Solution:

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

∴ log is not defined.

Example: 3: Find the value of ${{\log }_{\left( se{{c}^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ \right)}}1$

Solution:

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

Here, base = 1, ∴ log is not defined.

Example: 4: Find the value of ${{\log }_{30}}1$

Solution:

${{\log }_{30}}1=0$

Characteristic and Mantissa

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

In short, -3+0.951823 is written as $\overline{3}.9518231.$

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 decimal part of N (here, 0 < N < 1).
• If N > 1, the characteristic of log N will be on less than the number of digits in integral part of N.
• If 0 < N < 1, the characteristic of log N is negative and numerically it is one greater than the number of zeroes immediately after the decimal part in N.

For Example:

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

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

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

2. If $\log\;0.0000279=\overline{5}.4456042$

Here, four zeroes immediately after the decimal point in the number 0.0000279 is $\left( \overline{4+1} \right),i.e.\,\overline{5}.$

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 {log2+log3}

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

Following are the logarithm rules.

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

1. Loga (m n) = loga m + loga n

In general, loga (x1, x2, x3,…, xn) =

${{\log }_{a}}{{x}_{1}} +{{\log }_{a}}{{x}_{2}}+{{\log }_{a}}{{x}_{3}}+…+{{\log }_{a}}{{x}_{n}}$

$\left( where,\,{{x}_{1}},{{x}_{2}},{{x}_{3}},…,{{x}_{n}}>0 \right)$

Or

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

where, i = 1, 2, 3, …, n.

2. ${{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n$

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. ${{\log }_{b}}a.{{\log }_{a}}b=1\,\,\,\,\,\Leftrightarrow \,\,\,\,\,{{\log }_{b}}a=\frac{1}{{{\log }_{a}}b}$

7. ${{\log }_{b}}a.{{\log }_{c}}b.{{\log }_{a}}c=1$

8. ${{\log }_{y}}x.{{\log }_{z}}y.{{\log }_{a}}z={{\log }_{a}}x$

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

Some Additional Logarithm Properties

Logarithm properties are given below.

1. ${{a}^{{{\log }_{b}}x}}={{x}^{{{\log }_{b}}a}},b\ne 1,a,b,x$ are positive numbers.
2. ${{a}^{{{\log }_{a}}x}}=x,a>0,a\ne 1,x>0$
3. ${{\log }_{{{a}^{k}}}}x=\frac{1}{k}{{\log }_{a}}x,a>0,a\ne 1,x>0$
4. ${{\log }_{a}}{{x}^{2k}}=2k{{\log }_{a}}\left| x \right|,a>0,a\ne 1,k\in I$
5. ${{\log }_{{{a}^{^{2k}}}}}x=\frac{1}{2k}{{\log }_{\left| a \right|}}x,x>0,a\ne \pm 1\,\,and\,\,k\in I\tilde{\ }\left\{ 10 \right\}$
6. ${{\log }_{{{a}^{\alpha }}}}{{x}^{\beta }}=\frac{\beta }{\alpha }{{\log }_{a}}x,x>0,a>0,a\ne 1,\alpha \ne 0$
7. ${{\log }_{a}}{{x}^{2}}\ne 2{{\log }_{a}}x,a>0,a\ne 1$

Since, domain of ${{\log }_{a}}{{\left( x \right)}^{2}}$ is R ~ {0} and domain of ${{\log }_{a}}x\,\,is\,\,\left( 0,\infty \right)$ are not same.

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

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

Solution:

⇒ $3.x^{\log_{5}\;2} + 2^{\log_{5}\;x}=64$

⇒ $3.2^{\log_{5}\;x} + 2^{\log_{5}\;x}=64$ [by extra property (i)]

⇒ $4.2^{\log_5x} =64$

⇒ $2^{\log _{5}x}=4^2=2^4$

∴ $\log _{5}x=4$

∴ $x=5^4=625$

Example 2: If ${{4}^{{{\log }_{16\,}}4}}+{{9}^{{{\log }_{3}}9}}={{10}^{{{\log }_{x}}83}},$ find x.

Solution:

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

and ${{9}^{{{\log }_{3}}9}}={{9}^{2}}=81$ [by extra property (viii)]

∴ ${{4}^{{{\log }_{16}}4}}+{{9}^{{{\log }_{3}}9}}=2+81=83={{10}^{{{\log }_{x}}83}}$

⇒ ${{\log }_{10}}83={{\log }_{x}}83$

∴ x = 10

Example 3: Prove that ${{a}^{\sqrt{{{\log }_{a}}b}}}-{{b}^{\sqrt{{{\log }_{b}}a}}}=0.$

Solution:

Since, $\,\,\,\,\,{{a}^{\sqrt{\left( {{\log }_{a}}b \right)}}}={{a}^{\sqrt{{{\log }_{a}}b}\,\times \,\sqrt{{{\log }_{a}}b}\,\times \,\sqrt{{{\log }_{b}}a}}}$

= ${{a}^{{{\log }_{a}}b\,.\,\,\sqrt{{{\log }_{b}}a}}}$

= ${{b}^{\sqrt{{{\log }_{b}}a}}}$ [by extra property (ii)]

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

Example 4: Prove that $\frac{{{\log }_{2}}24}{{{\log }_{96}}2}-\frac{{{\log }_{2}}192}{{{\log }_{12}}2}=3.$

Solution:

$LHS=\frac{{{\log }_{2}}24}{{{\log }_{96}}2}-\frac{{{\log }_{2}}192}{{{\log }_{12}}2}$

= ${{\log }_{2}}24\times {{\log }_{2}}96-{{\log }_{2}}192\times {{\log }_{2}}12$

Now, let $12=\lambda ,$ then

$LHS={{\log }_{2}}2\lambda \times {{\log }_{2}}8\lambda -{{\log }_{2}}16\lambda \times {{\log }_{2}}\lambda$

= $\left( {{\log }_{2}}2+{{\log }_{2}}\lambda \right)\left( {{\log }_{2}}8+{{\log }_{2}}\lambda \right)$

$-\left( {{\log }_{2}}16+{{\log }_{2}}\lambda \right){{\log }_{2}}\lambda$

= $\left( {{\log }_{2}}2+{{\log }_{2}}\lambda \right)\left( {{\log }_{2}}{{2}^{3}}+{{\log }_{2}}\lambda \right)$

$-\left( {{\log }_{2}}{{2}^{4}}+{{\log }_{2}}\lambda \right){{\log }_{2}}\lambda$

= $\left( 1+{{\log }_{2}}\lambda \right)\left( 3{{\log }_{2}}2+{{\log }_{2}}\lambda \right)$

$-\left( 4{{\log }_{2}}2+{{\log }_{2}}\lambda \right){{\log }_{2}}\lambda$

= $\left( 1+{{\log }_{2}}\lambda \right)\left( 3+{{\log }_{2}}\lambda \right)-{{\log }_{2}}\lambda \left( 4+{{\log }_{2}}\lambda \right)$

= 3 = RHS

Properties Of Monotonocity Of Logarithm

Logarithm with Constant Base

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

Logarithm with Variable Base

1. logx a is defined, if $a>0,x>0,x\ne 1$
2. If a > 1, then logx a is monotonically decreasing in $\left( 0,1 \right)\cup \left( 1,\infty \right)$
3. If 0 < a < 1, then logx a is monotonically increasing in $\left( 0,1 \right)\cup \left( 1,\infty \right)$

Key Points

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

Graphs Of Logarithmic Functions

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

 

2. Graph of y = loga 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 = loga x, a > 1, x > 1
• y = loga x, 0 < a < 1, 0 < x < 1

 

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

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

3. Graph of $y={{\log }_{a}}\left| x \right|$

Graphs are symmetrical about Y-axis.

4. Graph of $y=\left| {{\log }_{a}} \right|\left. x \right\|$

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

5. Graph of $\left| y \right|=\left. {{\log }_{a}} \right|\left. x \right\|$

6. Graph of $y={{\log }_{a}}\left[ x \right],a>1\,and\,x\ge 1$

(where [ . ] denotes the greatest integer function)

Since, when $1\le x<2,\left[ x \right]=1\Rightarrow {{\log }_{a}}\left[ x \right]=0$

when $2\le x<3,\left[ x \right]=2\Rightarrow {{\log }_{a}}\left[ x \right]={{\log }_{a}}2$

when $3\le x<4,\left[ x \right]=3\Rightarrow {{\log }_{a}}\left[ x \right]={{\log }_{a}}3$ so on.

 

Important Shortcuts to Answer Logarithm Problems

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

Logarithm examples with solutions are given below.

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

Solution:

$\log {{10}^{11}}=11\log 10=11$

and log1110 = 10log11 = 10 × 1.0414 = 10.414

It is clear that, 11 > 10.414

⇒ $\log {{10}^{11}}>\log {{11}^{10}}$ [Since, base = 10]

⇒ ${{10}^{11}}>{{11}^{10}}$

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 ${{\log }_{2}}\left( x-2 \right)<{{\log }_{{{2}^{2}}}}\left( x-2 \right)=\frac{1}{2}{{\log }_{2}}\left( x-2 \right)$

⇒ ${{\log }_{2}}\left( x-2 \right)<\frac{1}{2}{{\log }_{2}}\left( x-2 \right)$

⇒ $\frac{1}{2}{{\log }_{2}}\left( x-2 \right)<0\Rightarrow {{\log }_{2}}\left( x-2 \right)<0$

⇒ $x-2<{{2}^{0}}$

⇒ x-2<1

⇒ x < 3 ……………… (ii)

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

$2<x<3\,\,or\,\,x\in \left( 2,3 \right)$

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

where, $a,b,c\in Q,$ the value of abc is

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

Solution:

Option: (c)

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

= ${{3}^{1/\left( 1-{{\log }_{4}}3 \right)}}={{3}^{1/{{\log }_{4}}\left( 4/3 \right)}}={{3}^{{{\log }_{4/3}}4}}$

∴ $a=3,b=\frac{4}{3},c=4$

Hence, $abc=3.\frac{4}{3}.4=16$

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

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

Solution: Option (a)

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

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

⇒ $\,\,\left( x-1 \right)\left( x-3 \right)>0$

⇒ x < 1 or x > 3 ……….(ii)

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

∴ x = 2, 3 ……….(iii)

From equations (ii) and (iii), use get $x\in \phi$

∴ Number of real roots = 0.

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

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

Solution:

Option: (c)

Since, $x={{\log }_{2a}}\left( \frac{bcd}{2} \right)$

⇒ $x+1={{\log }_{2a}}\left( \frac{2abcd}{2} \right)={{\log }_{2a}}\left( abcd \right)$

∴ $\frac{1}{x+1}={{\log }_{abcd}}2a$

Similarly, $\frac{1}{y+1}={{\log }_{abcd}}3b,\frac{1}{z+1}={{\log }_{abcd}}4c$

and $\frac{1}{w+1}={{\log }_{abcd}}5d$

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

= ${{\log }_{abcd}}\left( 120abcd \right)$

= ${{\log }_{abcd}}120+1$

= ${{\log }_{abcd}}N+1$ [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 (c) 1 (d) none of these

Solution:

Option: (c)

we have

$a={{\log }_{12}}18=\frac{{{\log }_{2}}18}{{{\log }_{2}}12}=\frac{1+2{{\log }_{2}}3}{2+{{\log }_{2}}3}$ and

$b={{\log }_{24}}54=\frac{{{\log }_{2}}54}{{{\log }_{2}}24}=\frac{1+3{{\log }_{2}}3}{3+{{\log }_{2}}3}$

Putting x = log2 3, we have

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

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

Problem 7: The value of $\frac{{{\log }_{2}}24}{{{\log }_{96}}2}-\frac{{{\log }_{2}}192}{{{\log }_{12}}2}$ is

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

Solution:

Option: (a)

Set log2 12 = a,

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

${{\log }_{2}}24=1+a,{{\log }_{2}}192={{\log }_{2}}\left( 16\times 12 \right)=4+a$

and $\frac{1}{{{\log }_{12}}2}={{\log }_{2}}12=a.$

Therefore, the given expression

$=\left( 1+a \right)\left( 3+a \right)-\left( 4+a \right)a=3$

Problem 8:  The solution of the equation ${{4}^{{{\log }_{2}}\log x}}=\log x-{{\left( \log x \right)}^{2}}+1$ is

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

Solution:

Option: (c)

log2 log x is meaningful if x > 1

Since ${{4}^{{{\log }_{2}}\log x}}={{2}^{2{{\log }_{2}}\log x}}={{\left( {{2}^{{{\log }_{2}}\log x}} \right)}^{2}}={{\left( \log x \right)}^{2}}$

$\left[ {{a}^{{{\log }_{a}}x}}=x,a>0,a\ne 1 \right]$

So the given equation reduces to

$2{{\left( \log x \right)}^{2}}-\log x-1=0$

⇒ $\log x=1,\log x=-1/2.$ But for x > 1

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

Problem 9:  If log0.5 (x – 1) < log0.25 (x – 1), then x lies in the interval.

• $\left( 2,\infty \right)$
• $\left( 3,\infty \right)$
• $\left( -\infty ,0 \right)$ (d) (0, 3)

Solution:

log0.5 (x – 1) < log0.25 (x – 1)

$\Leftrightarrow {{\log }_{0.5}}\left( x-1 \right)<{{\log }_{{{\left( 0.5 \right)}^{2}}}}\left( x-1 \right)$

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

$\Leftrightarrow \,\,\,\,{{\log }_{0.5}}\left( x-1 \right)<0$

$\Leftrightarrow \,\,\,\,\,\,\,x-1>1\Leftrightarrow x\in \left( 2,\infty \right)$

Problem 10:  If n = 2002 !, evaluate $\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+…+\frac{1}{{{\log }_{2002}}n}$

Solution:

We have, $\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+…+\frac{1}{{{\log }_{2002}}n}$

$={{\log }_{n}}2+{{\log }_{n}}3+{{\log }_{n}}4+…+{{\log }_{n}}2002$, Since, $\frac{1}{{{\log }_{b}}a}={{\log }_{a}}b$

= ${{\log }_{n}}\left( 2.3.4….2002 \right)$

= ${{\log }_{n}}\left( 2002! \right)={{\log }_{n}}n=1$

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 xx yy zz = 1

Solution:

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

⇒ $\log x=\lambda \left( y-z \right),\log y=\lambda \left( z-x \right),\log z=\lambda \left( x-y \right)$

⇒ $x\log x+y\log y+z\log z$

$=\lambda x\left( y-z \right)+\lambda y\left( z-x \right)+\lambda z\left( x-y \right)=0$

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

⇒ $\log \left( {{x}^{x}}{{y}^{y}}{{z}^{z}} \right)=0$

⇒ ${{x}^{x}}{{y}^{y}}{{z}^{z}}=1$

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

⇒ $x>1\,and\,{{\log }_{3}}\left( x-1 \right)>-\frac{5}{4}$

⇒ $x>1\,and\,x-1>{{3}^{-5/4}}$

⇒ $x>1\,and\,x>1+\frac{1}{{{3}^{5/4}}}$

⇒ $x>1+\frac{1}{{{3}^{5/4}}}$ ………. (i)

Now,

log3 {5 + 4 log3 (x – 1)} = 2

⇒ $5+4{{\log }_{3}}\left( x-1 \right)={{3}^{2}}$

⇒ ${{\log }_{3}}\left( x-1 \right)=1$

⇒ x – 1 = 3

∴ x=4

Clearly, x = 4 satisfies (i).

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

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

Solution: Clearly, the given equation is meaningful, if${{3}^{x}}-8>0\Rightarrow {{3}^{x}}>8\Rightarrow x>{{\log }_{3}}8$ ………. (i)

Now,

log3 (3x – 8) = 2 – x

⇒ ${{3}^{x}}-8=\frac{{{3}^{2}}}{{{3}^{x}}}$

⇒ ${{\left( {{3}^{x}} \right)}^{2}}-8\left( {{3}^{x}} \right)-9=0$

⇒ $\left( {{3}^{x}}-9 \right)\left( {{3}^{x}}+1 \right)=0$

⇒ ${{3}^{x}}-9=0\,\,\,\,\,\,\,\,\,\left[ Since, {{3}^{x}}>8\,\,Therefore, \,\,{{3}^{x}}+1\ne 0 \right]$

⇒ ${{3}^{x}}={{3}^{2}}$

∴ x = 2

Clearly, 2 > log3 8

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

Problem 14: Solve: x2 log x = 10x2

Solution:

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

Now,

x2 log x = 10x2

⇒ $\log \left\{ {{x}^{2\log x}} \right\}=\log \left( 10{{x}^{2}} \right)$

⇒ $2\log x.\log x=\log 10+\log {{x}^{2}}$

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

⇒ $2{{y}^{2}}-2y-1=0,\,where\,y=\log x$

⇒ $y=\frac{2\pm \sqrt{4+8}}{2}$

⇒ $y=1\pm \sqrt{3}$

⇒ ${{\log }_{10}}x=1\pm \sqrt{3}$

⇒ $x={{10}^{1\pm \sqrt{3}}}$

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

Solution:

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

9 – 2x > 0 and 3 – x > 0

⇒ $\,\,\,\,\,\,\,\,\,{{2}^{x}}<9\,and\,x<3$

⇒ $x<{{\log }_{2}}9\,and\,x<3$

x < 3 ………. (i)

Now,

log2 (9-2x) = 10log (3-x)

⇒ $9-{{2}^{x}}={{2}^{3-x}}$

⇒ $9-{{2}^{x}}=\frac{{{2}^{3}}}{{{2}^{x}}}$

$9\left( {{2}^{x}} \right)-{{\left( {{2}^{x}} \right)}^{2}}=8$

⇒ ${{\left( {{2}^{x}} \right)}^{2}}-9\left( {{2}^{x}} \right)+8=0$

⇒ ${{y}^{2}}-9y+8=0,\,where\,y={{2}^{x}}$

⇒ $\left( y-8 \right)\left( y-1 \right)=0$

⇒ y = 8, 1

⇒ ${{2}^{x}}=8,1$

⇒ x = 3, 0.

But, x = 3 does no satisfy (i).

Hence, x = 0.

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

Solution:

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

Now,

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

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

⇒ $\log \left( x+1 \right).\log \left( x+1 \right)=\log 100+\log \left( x+1 \right)$

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

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

∴ y = 2, -1

⇒ ${{\log }_{10}}\left( x+1 \right)=2,{{\log }_{10}}\left( x+1 \right)=-1$

⇒ $x+1={{10}^{2}},x+1={{10}^{-1}}$

⇒ x = 99, x = -0.9

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

Solution:

Let $x=\sqrt[3]{72.3}.$ Then,

log x = log(72.3)1/3

⇒ $\log x=\frac{1}{3}\log 72.3$

⇒ $\log x=\frac{1}{3}\times 1.8591$

⇒ $\log x=0.6197$

⇒ $\,\,\,\,\,\,\,\,\,x=anti\log \left( 0.6197 \right)$

⇒ x = 4.166

Problem 18:  Evaluate $\sqrt[5]{10076},$ if log 100.76 = 2.0029.

Solution:

Let $x=\sqrt[5]{10076}.$ Then,

log x =  log (10076)1/5

⇒ $\log x=\frac{1}{5}\log 10076$

⇒ $\log x=\frac{1}{5}\times 4.0029$

⇒ $\log x=0.8058$

⇒ $\,\,\,\,\,\,\,\,\,x=anti\log \left( 0.8058 \right)$

⇒ x = 6.409

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

Solution:

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

By using the formula ${{\log }_{a}}{{M}^{x}}=x{{\log }_{a}}.M\,and\,{{\log }_{{{a}^{x}}}}M=\frac{1}{x}{{\log }_{a}}M$ we can solve it.

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

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

Solution:

By solving we get $1.\overline{3}=\frac{4}{3},$ and use the formula ${{\log }_{a}}a=1.$

${{\log }_{4/3}}1.\overline{3}=1$

Let x = 1.333 ….. (i)

10x = 13.3333 ….. (ii)

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

So $9x=12\Rightarrow x=12/9,x=4/3;$

Now ${{\log }_{4/3}}\,1/\overline{3}={{\log }_{4/3}}\left( 4/3 \right)=1$

 

Problem 21: If

$N=n!\left( n\in N,n\ge 2 \right)\,then\,\underset{N\to \infty }{\mathop{\lim }}\,\left[ {{\left( {{\log }_{2}}N \right)}^{-1}}+{{\left( {{\log }_{3}}N \right)}^{-1}}+…+{{\left( {{\log }_{n}}N \right)}^{-1}} \right]$ is

Solution:

Here by using ${{\log }_{a}}b=\frac{1}{{{\log }_{b}}a}$ we can write given expansion as

${{\log }_{N}}2+{{\log }_{N}}3+……+{{\log }_{N}}n$ and then by using ${{\log }_{a}}\left( M.N \right)={{\log }_{a}}M+{{\log }_{a}}N$ and N = n!.

⇒ ${{\left( {{\log }_{2}}N \right)}^{-1}}+{{\left( {{\log }_{3}}N \right)}^{-1}}+……+{{\left( {{\log }_{n}}N \right)}^{-1}}={{\log }_{N}}2+{{\log }_{N}}3+……+{{\log }_{N}}n={{\log }_{n}}\left( 2.3….N \right)={{\log }_{N}}N=1.$

Problem 22:  If $\log {{x}^{2}}-\log 2x=3\log 3-\log 6$ then x equals

Solution:

By using ${{\log }_{a}}\left( M.N \right)={{\log }_{a}}M+{{\log }_{a}}N\,and\,{{\log }_{a}}{{M}^{x}}=x{{\log }_{a}}.M$

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

$2\log x-\log 2-\log x=3\log 3-\log 2-\log 3\Rightarrow \log x=2\log 3\Rightarrow x=9$

Problem 23:  Prove that, ${{\log }_{2-\sqrt{3}}}\left( 2+\sqrt{3} \right)=-1$

Solution:

By multiplying and dividing by $2+\sqrt{3}\,to\,2-\sqrt{3}$ we will get $2+\sqrt{3}=\frac{1}{2-\sqrt{3}}.$ Therefore by using ${{\log }_{1/N}}N=-1$ we can easily prove this.

⇒ ${{\log }_{2-\sqrt{3}}}\frac{1}{2-\sqrt{3}}\Rightarrow {{\log }_{2-\sqrt{3}}}{{\left( 2-\sqrt{3} \right)}^{-1}}\Rightarrow -1.{{\log }_{2-\sqrt{3}}}\left( 2-\sqrt{3} \right)=-1$

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

Solution:

Here $\sqrt{5\sqrt{5\sqrt{5……..\infty }}}$ can be represented as $y=\sqrt{5y}\,\,where\,\,y=\sqrt{5\sqrt{5\sqrt{5……..\infty }}}.$

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

Let $y=\sqrt{5\sqrt{5\sqrt{5……..\infty }}}$

⇒ $y=\sqrt{5y}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow {{y}^{2}}=5y\,\,or\,\,{{y}^{2}}-5y=0$

⇒ $y\left( y-5 \right)=0\,\,\,\,\,\,\,\Rightarrow y=0,y=5$

∴ $\,\,\,\,\,\,{{\log }_{5}}5=1$

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

Solution:

As similar to Example 3 we can solve it by using ${{\log }_{1/N}}N=-1.$

x = 0.4444 ….. (i)

10x = 4.4444 ….. (ii)

Equation (ii) – Equation (i)

So $9x=4\Rightarrow x=4/9$

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 ${{2}^{{{\log }_{6}}18}}{{.3}^{{{\log }_{6}}3}}$

Solution:

We can solve above problem by using ${{\log }_{a}}\left( M.N \right)={{\log }_{a}}M+{{\log }_{a}}N\,and\,\,{{a}^{{{\log }_{e}}c}}={{c}^{{{\log }_{e}}a}}$ step by step.

⇒

${{2}^{{{\log }_{6}}18}}{{\left( 3 \right)}^{{{\log }_{6}}3}}={{2}^{{{\log }_{6}}\left( 6\times 3 \right)}}{{.3}^{{{\log }_{6}}3}}={{2}^{1+{{\log }_{6}}3}}{{.3}^{{{\log }_{6}}3}}={{2.2}^{{{\log }_{6}}3}}{{.3}^{{{\log }_{6}}3}}\,\,\,\left( Since, {{a}^{{{\log }_{e}}c}}={{c}^{{{\log }_{e}}a}} \right)$

= $2.{{\left( 3 \right)}^{{{\log }_{6}}2}}.{{\left( 3 \right)}^{{{\log }_{6}}3}}=2{{\left( 3 \right)}^{{{\log }_{6}}2+{{\log }_{6}}3}}=2{{\left( 3 \right)}^{{{\log }_{6}}\left( 6 \right)}}=2.\left( 3 \right)=6$

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

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

Therefore by using formula $y={{\log }_{a}}x\Leftrightarrow {{a}^{y}}=x$ we can write $co{{s}^{3}}\alpha ={{\left( \sec \,\,\alpha \right)}^{x}}.$ Hence by solving this we will 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 $k\,\in \,N,$ such that ${{\log }_{2}}x+{{\log }_{4}}x+{{\log }_{8}}x={{\log }_{k}}x\,and\,\,\forall x\in R’\,\,If\,\,k={{\left( a \right)}^{1/b}}$ then find the value of $a+b;a\in N,b\in N$ and b is a prime number.

Solution:

By using ${{\log }_{b}}a=\frac{{{\log }_{c}}a}{{{\log }_{c}}b}=\frac{\log a}{\log b}$

We can obtain the value of k and then by comparing it to $k={{\left( a \right)}^{1/b}}$ we can obtain value of a + b.

Given, $\frac{\log x}{\log 2}+\frac{\log x}{2\log 2}+\frac{\log x}{3\log 2}=\frac{\log x}{\log k}\Rightarrow \frac{\log x}{\log 2}\left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3} \right]=\frac{\log x}{\log 2}\left( \frac{11}{6} \right)=\frac{\log x}{\log k}\Rightarrow \log x\left[ \frac{11}{6}\frac{1}{\log 2}-\frac{1}{\log k} \right]=0$

Also, $\frac{11}{6}\frac{1}{\log 2}-\frac{1}{\log k}=0\Rightarrow \frac{11}{6}=\frac{\log 2}{\log k}\Rightarrow \frac{11}{6}={{\log }_{k}}2$

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

Comparing by $k={{\left( a \right)}^{1/b}}\Rightarrow a=64\,\,and\,\,b=11\Rightarrow a+b=64+11=75$

Problem 29: ${{\log}_{e}}\,[{{(1+x)}^{1+x}}{{(1-x)}^{1-x}}]\,=$

Solution:

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

Problem 30: In the expansion of $2{{\log}_{e}}x-{{\log}_{e}}(x+1)-{{\log}_{e}}(x-1)$, the coefficient of ${{x}^{-4}}$ is

Solution:

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

Problem 31: ${{\log}_{e}}2+{{\log}_{e}}\left(1+\frac{1}{2} \right)+{{\log}_{e}}\left(1+\frac{1}{3} \right)+….+{{\log}_{e}}\left(1+\frac{1}{n-1} \right)=$

Solution:

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

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

Solution:

$\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 \ So, \ coefficient \ of \ {{x}^{n}}={{(-1)}^{n-1}}\left(\frac{{{2}^{n}}+1}{n} \right)\\$$[\,\,{{(-1)}^{n}}={{(-1)}^{n+2}}=…]\\$

Problem 33: $1+\left(\frac{1}{2}+\frac{1}{3} \right)\,\frac{1}{4}+\left(\frac{1}{4}+\frac{1}{5} \right)\,\frac{1}{{{4}^{2}}}+\left(\frac{1}{6}+\frac{1}{7} \right)\,\frac{1}{{{4}^{3}}}+….\infty =$

Solution:

$S=\left\{1+\frac{{{\left(\frac{1}{2} \right)}^{2}}}{2}+\frac{{{\left(\frac{1}{2} \right)}^{4}}}{4}+….. \right\}+2\left\{\frac{1}{2}+\frac{{{\left( \frac{1}{2} \right)}^{3}}}{3}+\frac{{{\left( \frac{1}{2} \right)}^{5}}}{5\ }+….. \right\}-1\\ =1-\frac{1}{2}{{\log }_{e}}\left(1+\frac{1}{2} \right)\text{ }\left(1-\frac{1}{2} \right)+{{\log }_{e}}\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}} \right)-1\\ =-\frac{1}{2}{{\log}_{e}}\frac{3}{4}+{{\log}_{e}}3\\={{\log}_{e}}2\sqrt{3}.$