# Logarithm

## Introduction to Logarithm

The logarithm of any positive number, whose base is a number (> 0) different from 1, is the index or the power to which the base must be raised in order to obtain the given number.

i.e. 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, $\left[\ln \,b={{\log }_{10}}b\times {{\log }_{e}}10=\frac{1}{{{\log }_{10}}e}\times {{\log }_{10}}b \right.\left. =\frac{1}{0.4343}{{\log }_{10}}b=2303{{\log }_{10}}b \right]$

⇒ 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. $\,\,\,\,\,{{\log }_{a}}0=-8\left( if\,a>1 \right)$
2. If 0 < a < 1, then ${{u}^{\infty }}=0$ 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$=\tan 45{}^\circ =1\tan$.

∴ log is not defined.

Example: 2: Find the value of ${{\log }_{\left( se{{c}^{2}}60{}^\circ -{{\tan }^{2}}60{}^\circ \right)}}\cos 60{}^\circ$

Solution:

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

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

For Example:

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

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

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

2. 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

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) = loga x1

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

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

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

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

Solution:

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

⇒ $3.2^{\log_{5}\;x} + 2^{\log_{5}\;x}=64$ [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 $y={{\log }_{a}}\left| x \right|$

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

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:

(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:

${{3}^{{{\log }_{3}}\left( {{x}^{2}}-4x+3 \right)}}=\left( x-3 \right)$ ……….(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 log6 a+ log6 b+ log6 c = 6, where $a,b,c\in N$ and a, b, c are in GP and b – a is a square of an integer, then the value of a + b – c is

(a) 21 (b) 15 (c) 9 (d) 3

Solution:

(b) Since, log6 a+ log6 b+ log6 c = 6

⇒ ${{\log }_{6}}\left( abc \right)=6$

∴ abc =6

${{b}^{3}}\,={{6}^{6}}\,\,\,\,\,\,\left[ Since, {{b}^{2}}=ac \right]$

⇒ b = 36

Problem 6:  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:

(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]

Hence, N = 120

Problem 7:  If a = log12 18, b = log24 54 then the value of ab + (a–b) is

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

Solution:

(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 8: 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:

(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 9:  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:

(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 10:  If log x2 – log 2 x = 3 log 3 – log 6 then x equals

(a) 9 (b) 3 (c) 4 (d) 5

Solution:

(a) Clearly x > 0. The given equation can be written as

2 log x – log 2 – log x = 3 log 3 – log 2 – log 3

⇒ $\log x=2,\log 3\Rightarrow x=9.$

Problem 11:  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 12:  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 13:  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 14:  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 15: 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

⇒ $\left( {{3}^{x}}-8 \right)=2-x$

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

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

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

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

⇒ ${{3}^{x}}={{3}^{2}}$

∴ x = 2

Clearly, 2 > log3 8

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

Problem 16: 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 17:  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)

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

⇒ $9-{{2}^{x}}={{2}^{3-x}}$

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

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

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

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

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

⇒ y = 8, 1

⇒ ${{2}^{x}}=8,1$

⇒ x = 3, 0.

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

Hence, x = 0.

Problem 18:  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 19:  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 20:  Evaluate $\sqrt[5]{10076},$ if log 100.76 = 2.0029.

Solution:

Let $x=\sqrt[5]{10076}.$ Then,

logx 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 21:  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 ${{\log }_{a}}{{M}^{x}}=x{{\log }_{a}}.M\,and\,{{\log }_{{{a}^{x}}}}M=\frac{1}{d}{{\log }_{a}}M$ 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 22:  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 23:  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 24:  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 25:  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: 26 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 27 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, $2.25=\frac{225}{100}=\frac{9}{4};\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\log }_{2.25}}\left( 0.\overline{4} \right)={{\log }_{\left( \frac{9}{4} \right)}}\left( \frac{4}{9} \right)=-1$

Problem28:  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 29:  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 suing 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 30:  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 $k={{\left( a \right)}^{1/b}}\Rightarrow a=64\,\,and\,\,b=11\Rightarrow a+b=64+11=75$