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# Logarithm

## Introduction toΒ 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 ax = b (where a > 0, β  1), then x is called the logarithm of b to the base a, and we write loga b = x, clearly b > 0. Thus,

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

If a = 10, then we write log b rather than log10 b. If a = e, we write ln b rather than loge b. Here, βeβ is called Napierβs base and has a numerical value equal to 2.7182. Also, log10 e is known as the Napierian constant.

i.e., log10 e = 0.4343

β΄ ln b = 2.303 log10 b

$$\begin{array}{l}\left[\text{since}\ \ln \,b={{\log }_{10}}b\times {{\log }_{e}}10=\frac{1}{{{\log }_{10}}e}\times {{\log }_{10}}b \right.\left. =\frac{1}{0.4343}{{\log }_{10}}b=2.303\ {{\log }_{10}}b \right]\end{array}$$

β 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:Β From the definition of the logarithm of the number b to the base a, we have the identity

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

which is known as the Fundamental Logarithmic Identity.

Corollary 2:Β The function defined by

$$\begin{array}{l}f\left( x \right)={{\log }_{a}}x,a>0,a\ne 1\end{array}$$
is called logarithmic function. Domain is (0, β), and the range is R (set of all real numbers).

Corollary 3: ax > 0, β x β R

1. If a > 1, then ax is monotonically increasing. For example,
$$\begin{array}{l}{{5}^{2.7}}>{{5}^{2.5}},{{3}^{222}}>{{3}^{111}}\end{array}$$
2. 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 aβ = 0. i.e., loga 0 = –β (if a > 1)Β
2. If 0 < a < 1, then aβ = 0. i.e., loga 0 = +β (if 0 < a < 1)Β

Corollary 5:

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

Remarks

1. βlogβ is the abbreviation of the word βlogarithmβ.
2. Common logarithm (Briggβs logarithms). The base is 10.
3. If x < 0, a > 0 and a β  1,Β then loga x is imaginary.
4. $$\begin{array}{l}{{\log }_{a}}1=0\left( a>0,a\ne 1 \right)\end{array}$$
5. $$\begin{array}{l}{{\log }_{a}}a=1\left( a>0,a\ne 1 \right)\end{array}$$
6. $$\begin{array}{l}{{\log }_{\left( 1/a \right)}}a=-1\left( a>0,a\ne 1 \right)\end{array}$$

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

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

### Solved Examples on Logarithm

Example: 1:

$$\begin{array}{l}\text{Find the value of }\ {{\log }_{\tan 45{}^\circ }}\cot 30{}^\circ\end{array}$$

Solution:

Here, baseΒ tan 450 = 1

β΄ the log is not defined.

Example: 2:

$$\begin{array}{l}\text{Find the value of }\ log_{(sec^2 60^0 – tan^2 60^0)} cos 60^0\end{array}$$

Solution:

Here, base

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

β΄ the log is not defined.

Example: 3:

$$\begin{array}{l}\text{Find the value of }\ {{\log }_{\left( se{{c}^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ \right)}}1\end{array}$$

Solution:

Since,

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

Here, base = 1, β΄ the log is not defined.

Example: 4:

$$\begin{array}{l}\text{Find the value of }\ {{\log }_{30}}1\end{array}$$

Solution:

$$\begin{array}{l}{{\log }_{30}}1=0\end{array}$$

## Characteristic and Mantissa

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

i.e., log N = Integer + Fractional or decimal part (+ve)

β 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 β 0.0481769 = (-2 – 1) + (1 – 0.0481769) = -3 + 0.9518231

Hence, -3 is the characteristic and 0.9518231

(Not 0.0481769) is mantissa of log 0.00895.

$$\begin{array}{l}\text{In short, -3 + 0.951823 is written as}\ \overline{3}.9518231.\end{array}$$

### Important Conclusions on Characteristic and Mantissa

• If the characteristics of log N be n, then the number of digits in N is (n+1) (Here, N > 1).
• If the characteristics of log N be βn, then there exists (n-1) number of zeroes after the decimal part of N (here, 0 < N < 1).
• If N > 1, the characteristic of log N will be less than the number of digits in an integral part of N.
• If 0 < N < 1, the characteristic of log N is negative, and numerically, it is one greater than the number of zeroes immediately after the decimal part in N.

For example:

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

β΄ Number of digits in an integral part of N = 3

β Characteristic of log 235.68 = N -1 = 3 β 1 = 2

2.

$$\begin{array}{l}\text{If}\ \log\;0.0000279=\overline{5}.4456042\end{array}$$

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

$$\begin{array}{l}\left( \overline{4+1} \right),i.e.\,\overline{5}.\end{array}$$

Problem:Β If log 2 = 0.301 and log 3 = 0.477, find the number of digits in 620.

Solution: Let P = 620 = (2Γ3)20

β΄ log P = 20 log (2Γ3) = 20 {log 2+ log 3}

= 20 {0.301 + 0.477} = 20 Γ 0.778 = 15.560

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

## Principle Properties of Logarithm

Following are the logarithm rules:

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

1. Loga (m n) = loga m + loga n
In general, loga (x1, x2, x3,β¦, xn) = loga x1 + loga x2 + loga x3 +β¦ + loga xn (where x1, x2, x3,β¦, xn > 0)
Or
$$\begin{array}{l}{{\log }_{a}}\left( \prod\limits_{i=1}^{n}{{{x}_{i}}} \right)=\sum\limits_{i=1}^{n}{{{\log }_{a}}{{x}_{i}},\forall {{x}_{i}}}>0\end{array}$$
where, i = 1, 2, 3, β¦, n.
2. $$\begin{array}{l}{{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n\end{array}$$
3. $$\begin{array}{l}{{\log }_{a}}{{m}^{\alpha }}=\alpha {{\log }_{a}}m\end{array}$$
4. $$\begin{array}{l}{{\log }_{{{a}^{\beta }}}}m=\frac{1}{\beta }{{\log }_{a}}m\end{array}$$
5. $$\begin{array}{l}{{\log }_{b}}m=\frac{{{\log }_{a}}m}{{{\log }_{a}}b}\end{array}$$
6. $$\begin{array}{l}{{\log }_{b}}a.{{\log }_{a}}b=1\,\,\,\,\,\Leftrightarrow \,\,\,\,\,{{\log }_{b}}a=\frac{1}{{{\log }_{a}}b}\end{array}$$
7. $$\begin{array}{l}{{\log }_{b}}a.{{\log }_{c}}b.{{\log }_{a}}c=1\end{array}$$
8. $$\begin{array}{l}{{\log }_{y}}x.{{\log }_{z}}y.{{\log }_{a}}z={{\log }_{a}}x\end{array}$$
9. $$\begin{array}{l}{{e}^{\ln \,{{a}^{x}}}}={{a}^{x}}\end{array}$$

Logarithm properties are given below.

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. $$\begin{array}{l}{{\log }_{{{a}^{k}}}}x=\frac{1}{k}{{\log }_{a}}x,a>0,a\ne 1,x>0\end{array}$$
4. $$\begin{array}{l}{{\log }_{a}}{{x}^{2k}}=2k{{\log }_{a}}\left| x \right|,a>0,a\ne 1,k\in I\end{array}$$
5. $$\begin{array}{l}{{\log }_{{{a}^{^{2k}}}}}x=\frac{1}{2k}{{\log }_{\left| a \right|}}x,x>0,a\ne \pm 1\,\,and\,\,k\in I\tilde{\ }\left\{ 10 \right\}\end{array}$$
6. $$\begin{array}{l}{{\log }_{{{a}^{\alpha }}}}{{x}^{\beta }}=\frac{\beta }{\alpha }{{\log }_{a}}x,x>0,a>0,a\ne 1,\alpha \ne 0\end{array}$$
7. $$\begin{array}{l}{{\log }_{a}}{{x}^{2}}\ne 2{{\log }_{a}}x,a>0,a\ne 1\end{array}$$
The domain of loga (x)2 is R ~ {0}, and the domain of loga x is (0, β) are not the same.
8. $$\begin{array}{l}a_{b}^{log\;a} =\sqrt{a},\,if\,b={a}^{2},a>0,b>0,b\ne 1\end{array}$$
9. $$\begin{array}{l}a_{b}^{log\;a} =a^{2},\,if\,b=\sqrt{a},a>0,b>0,b\ne 1\end{array}$$

Example 1:Β Solve the equation

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

Solution:

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

$$\begin{array}{l}\Rightarrow 3.2^{\log_{5}\;x} + 2^{\log_{5}\;x}=64\end{array}$$
[by extra property (i)]

$$\begin{array}{l}\Rightarrow 4.2^{\log_5x} =64\end{array}$$

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

$$\begin{array}{l}\therefore \log _{5}x=4\end{array}$$

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

Example 2:

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

Solution:

Since,

$$\begin{array}{l}{{4}^{{{\log }_{16}}4}}=\sqrt{4}=2\end{array}$$
[by extra property (ix)]

and

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

$$\begin{array}{l}\therefore {{4}^{{{\log }_{16}}4}}+{{9}^{{{\log }_{3}}9}}=2+81=83={{10}^{{{\log }_{x}}83}}\end{array}$$

$$\begin{array}{l}\Rightarrow {{\log }_{10}}83={{\log }_{x}}83\end{array}$$

β΄ x = 10

Example 3:

$$\begin{array}{l}\text{Prove that}\ {{a}^{\sqrt{{{\log }_{a}}b}}}-{{b}^{\sqrt{{{\log }_{b}}a}}}=0.\end{array}$$

Solution:

Since,

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

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

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

Hence,

$$\begin{array}{l}{{a}^{\sqrt{\left( {{\log }_{a}}b \right)}}}-{{b}^{\sqrt{\left( {{\log }_{b}}a \right)}}}=0\end{array}$$

Example 4:

$$\begin{array}{l}\text{Prove that}\ \frac{{{\log }_{2}}24}{{{\log }_{96}}2}-\frac{{{\log }_{2}}192}{{{\log }_{12}}2}=3.\end{array}$$

Solution:

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

Β

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

Now, let 12 = Ξ»,Β then

$$\begin{array}{l}LHS={{\log }_{2}}2\lambda \times {{\log }_{2}}8\lambda -{{\log }_{2}}16\lambda \times {{\log }_{2}}\lambda\end{array}$$

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

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

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

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

= 3

= RHS

## Properties of Monotonocity of Logarithm

### Logarithm with Constant Base

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

### Logarithm with Variable Base

1. logx a is defined, if a > 0, x > 0, x β  1
2. If a > 1, then logx a is monotonically decreasing in (0, 1) U (1, β)
3. If 0 < a < 1, then logx a is monotonically increasing in (0, 1) U (1, β)

### 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 βaβ 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

$$\begin{array}{l}y={{\log }_{a}}\left| x \right|\end{array}$$

4. Graph of

$$\begin{array}{l}y=\left| {{\log }_{a}} \right|\left. x \right\|\end{array}$$

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

5. Graph of

$$\begin{array}{l}\left| y \right|=\left. {{\log }_{a}} \right|\left. x \right\|\end{array}$$

6. Graph of

$$\begin{array}{l}y={{\log }_{a}}\left[ x \right],a>1\,and\,x\ge 1\end{array}$$

(Where [ . ] denotes the greatest integer function)

Since, when 1 β€ x < 2, [x] = 1 β loga [x] = 0

When 2 β€ x < 3, [x] = 2 β loga [x] = loga 2

When 3 β€ x < 4, [x] = 3 β loga [x] = loga 3 andΒ so on.

### Important Shortcuts to Answer Logarithm Problems

1. For a non-negative number βaβ and
$$\begin{array}{l}n\ge 2,n\in N,\sqrt[n]{a}={{a}^{1/n}}.\end{array}$$
2. The number of positive integers having base a and characteristic n is
$$\begin{array}{l}{{a}^{n+1}}-{{a}^{n}}.\end{array}$$
3. The logarithm of zero and negative real numbers is not defined.
4. $$\begin{array}{l}\left| {{\log }_{b}}a+{{\log }_{a}}b \right|\ge 2,\forall a>0,a\ne 1,b>0,b\ne 1\end{array}$$
5. $$\begin{array}{l}{{\log }_{2}}{{\log }_{2}}\underbrace{\sqrt{\sqrt{\sqrt{\sqrt{…\sqrt{2}}}}}}_{n\,\,times}=-n\end{array}$$
6. $$\begin{array}{l}{{a}^{\sqrt{{{\log }_{a}}b}}}={{b}^{\sqrt{{{\log }_{b}}a}}}\end{array}$$
7. Logarithms to the base 10 are called common logarithms (Briggβs logarithms).
8. If
$$\begin{array}{l}x={{\log }_{c}}b+{{\log }_{b}}c,y={{\log }_{a}}c+{{\log }_{c}}a,z={{\log }_{a}}b+{{\log }_{b}}a,\,\,then\,\,{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4=xyz.\end{array}$$

## Practice Problems onΒ Logarithm

Logarithm examples with solutions are given below.

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

Solution:

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

and log1110 = 10log11 = 10 Γ 1.0414 = 10.414

It is clear that 11 > 10.414

$$\begin{array}{l}\Rightarrow \log {{10}^{11}}>\log {{11}^{10}}\end{array}$$
[Since, base = 10]

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

Problem 2: If log2 (x – 2) < log4 (x – 2), find the interval in which x lies.

Solution:

Here, x β 2 > 0

β x > 2 β¦β¦β¦β¦β¦β¦ (i)

and

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

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

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

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

β x – 2 < 1

β x < 3 β¦β¦β¦β¦β¦β¦ (ii)

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

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

Problem 3:Β If

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

where, a, b, c β Q, the value of abc is

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

Solution:

Option: (c)

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

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

β΄ a = 3, b = 4/3 and c = 4

Hence,

$$\begin{array}{l}abc=3.\frac{4}{3}.4=16\end{array}$$

Problem 4:Β Number of real roots of equation

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

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

Solution: Option (a)

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

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

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

β x < 1 or x > 3 β¦β¦β¦.(ii)

Equation (i) reduces to

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

β΄ x = 2, 3 β¦β¦β¦.(iii)

From equations (ii) and (iii), we get x β Ξ¦.

β΄ The number of real roots = 0

Problem 5:Β  If

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

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

Solution:

Option: (c)

Since,

$$\begin{array}{l}x={{\log }_{2a}}\left( \frac{bcd}{2} \right)\end{array}$$

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

$$\begin{array}{l}\therefore\frac{1}{x+1}={{\log }_{abcd}}2a\end{array}$$

Similarly,

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

and

$$\begin{array}{l}\frac{1}{w+1}={{\log }_{abcd}}5d\end{array}$$

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

Β

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

Β

$$\begin{array}{l}={{\log }_{abcd}}120+1\end{array}$$

$$\begin{array}{l}={{\log }_{abcd}}N+1\end{array}$$
[given]

On comparing, we have Β N = 120

Problem 6:Β  If a = log12 18, b = log24 54, then the value of ab +5 (aβb) is

(a) 0

(b) 4

(c) 1

(d) None of the above

Solution:

Option: (c)

We have

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

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

Putting x = log2 3, we have

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

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

Problem 7:

$$\begin{array}{l}\text{The value of }\ \frac{{{\log }_{2}}24}{{{\log }_{96}}2}-\frac{{{\log }_{2}}192}{{{\log }_{12}}2}\ \text{is}\end{array}$$

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

Solution:

Option: (a)

Set log2 12 = a,

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

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

and

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

Therefore, the given expression

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

Problem 8:Β  The solution of the equation

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

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

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

So the given equation reduces to

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

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

But for x > 1,

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

Problem 9:Β  If log0.5 (x β 1) < log0.25 (x β 1), then x lies in the interval.

(a) (2, β)

(b) (3, β)

(c) (-β, 0)

(d) (0, 3)

Solution:

log0.5 (x β 1) < log0.25 (x β 1)

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

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

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

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

Problem 10:Β  If n = 2002 !, evaluate

$$\begin{array}{l}\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+…+\frac{1}{{{\log }_{2002}}n}\end{array}$$

Solution:

We have,

$$\begin{array}{l}\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+…+\frac{1}{{{\log }_{2002}}n}\end{array}$$

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

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

$$\begin{array}{l}={{\log }_{n}}\left( 2002! \right)={{\log }_{n}}n=1\end{array}$$

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

$$\begin{array}{l}\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y},\end{array}$$
prove that xx yy zz = 1.

Solution:

Let

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

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

Β

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

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

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

Β

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

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

Problem 12:Β  Solve: log3 {5 + 4 log3 (x β 1)} = 2

Solution:

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

$$\begin{array}{l}\Rightarrow x>1\,and\,{{\log }_{3}}\left( x-1 \right)>-\frac{5}{4}\end{array}$$

$$\begin{array}{l}\Rightarrow x>1\,and\,x-1>{{3}^{-5/4}}\end{array}$$

$$\begin{array}{l}\Rightarrow x>1\,and\,x>1+\frac{1}{{{3}^{5/4}}}\end{array}$$

Β

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

Now,

log3 {5 + 4 log3 (x β 1)} = 2

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

Β

$$\begin{array}{l}\Rightarrow {{\log }_{3}}\left( x-1 \right)=1\end{array}$$

β x – 1 = 3

β΄ x=4

Clearly, x = 4 satisfies (i).

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

Problem 13:Β Solve log3 (3x – 8) = 2 β x

Solution: Clearly, the given equation is meaningful, if

$$\begin{array}{l}{{3}^{x}}-8>0\Rightarrow {{3}^{x}}>8\Rightarrow x>{{\log }_{3}}8…..(i)\end{array}$$

Now,

log3 (3x – 8) = 2 β x

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

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

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

$$\begin{array}{l}\Rightarrow {{3}^{x}}-9=0\,\,\,\,\,\,\,\,\,\left[ Since, {{3}^{x}}>8\,\,Therefore, \,\,{{3}^{x}}+1\ne 0 \right]\end{array}$$

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

β΄ x = 2

Clearly, 2 > log3 8

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

Problem 14:Β Solve: x2 log x = 10x2

Solution:

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

Now,

x2 log x = 10x2

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

Β

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

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

Β

$$\begin{array}{l}\Rightarrow 2{{y}^{2}}-2y-1=0,\,where\,y=\log x\end{array}$$

Β

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

Β

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

$$\begin{array}{l}\Rightarrow {{\log }_{10}}x=1\pm \sqrt{3}\end{array}$$

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

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

Solution:

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

9 β 2x > 0 and 3 β x > 0

Β

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

$$\begin{array}{l}\Rightarrow x<{{\log }_{2}}9\ \text{and}\ x<3\end{array}$$

x < 3 β¦β¦β¦. (i)

Now,

log2 (9-2x) = 10log (3-x)

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

Β

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

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

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

Β

$$\begin{array}{l}\Rightarrow {{y}^{2}}-9y+8=0,\,where\,y={{2}^{x}}\end{array}$$

Β

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

β y = 8, 1

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

β x = 3, 0

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

Hence, x = 0

Problem 16:Β  Solve:

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

Solution:

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

Now,

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

Β

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

Β

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

Β

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

$$\begin{array}{l}\Rightarrow {{y}^{2}}-y-2=0,\,where\,y=\log \left( x+1 \right)\end{array}$$

β΄ y = 2, -1

$$\begin{array}{l}\Rightarrow {{\log }_{10}}\left( x+1 \right)=2,{{\log }_{10}}\left( x+1 \right)=-1\end{array}$$

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

β x = 99, x = -0.9

Problem 17:Β  Evaluate

$$\begin{array}{l}\sqrt[3]{72.3},\,\,if\,\,\log 0.723=\overline{1}.8591.\end{array}$$

Solution:

Let

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

log x = log(72.3)1/3

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

Β

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

Β

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

$$\begin{array}{l}\Rightarrow x=anti\log \left( 0.6197 \right)\end{array}$$

β x = 4.166

Problem 18:Β

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

Solution:

Let

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

log x =Β  log (10076)1/5

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

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

Β

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

$$\begin{array}{l}\Rightarrow x=anti\log \left( 0.8058 \right)\end{array}$$

β x = 6.409

Problem 19:Β  What is logarithm of

$$\begin{array}{l}32\sqrt[5]{4}\,\,to\,the\,base\,\,2\sqrt{2}\end{array}$$

Solution:

Here we can write

$$\begin{array}{l}32\sqrt[5]{4}\,\,as\,\,{{2}^{5}}{{4}^{1/5}}={{\left( 2 \right)}^{27/5}}\,and\,2\sqrt{2}\,\,as\,{{2}^{\frac{3}{2}}}\end{array}$$

By using the formula

$$\begin{array}{l}{{\log }_{a}}{{M}^{x}}=x{{\log }_{a}}.M\,and\,{{\log }_{{{a}^{x}}}}M=\frac{1}{x}{{\log }_{a}}M\end{array}$$
we can solve it.

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

Problem 20:Β  Prove that,

$$\begin{array}{l}{{\log }_{4/3}}\left( 1.\overline{3} \right)=1\end{array}$$

Solution:

By solving we get

$$\begin{array}{l}1.\overline{3}=\frac{4}{3},\end{array}$$
and use the formula
$$\begin{array}{l}{{\log }_{a}}a=1.\end{array}$$

$$\begin{array}{l}{{\log }_{4/3}}1.\overline{3}=1\end{array}$$

Let x = 1.333 β¦.. (i)

10x = 13.3333 β¦.. (ii)

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

So

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

Now

$$\begin{array}{l}{{\log }_{4/3}}\,1/\overline{3}={{\log }_{4/3}}\left( 4/3 \right)=1\end{array}$$

Problem 21:Β If

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

Solution:

Here, by using

$$\begin{array}{l}{{\log }_{a}}b=\frac{1}{{{\log }_{b}}a}\end{array}$$
we can write given expansion as

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

$$\begin{array}{l}\Rightarrow {{\left( {{\log }_{2}}N \right)}^{-1}}+{{\left( {{\log }_{3}}N \right)}^{-1}}+……+{{\left( {{\log }_{n}}N \right)}^{-1}}\\={{\log }_{N}}2+{{\log }_{N}}3+……+{{\log }_{N}}n={{\log }_{n}}\left( 2.3….N \right)={{\log }_{N}}N=1.\end{array}$$

Problem 22:Β

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

Solution:

By using

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

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

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

Problem 23:Β  Prove that,

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

Solution:

By multiplying and dividing by 2 + β3 to 2 – β3, we will get

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

Therefore, by using

$$\begin{array}{l}{{\log }_{1/N}}N=-1\end{array}$$
we can easily prove this.

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

Example 24:Β Prove that,

$$\begin{array}{l}{{\log }_{5}}\sqrt{5\sqrt{5\sqrt{5……..\infty }}}=1\end{array}$$

Solution:

Here,

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

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

Let

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

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

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

Β

$$\begin{array}{l}\therefore {{\log }_{5}}5=1\end{array}$$

Problem 25:Β Prove that,

$$\begin{array}{l}{{\log }_{2.25}}\left( 0.\overline{4} \right)=-1\end{array}$$

Solution:

Similar to example 3, we can solve it by using

$$\begin{array}{l}{{\log }_{1/N}}N=-1.\end{array}$$

x = 0.4444 β¦.. (i)

10x = 4.4444 β¦.. (ii)

Equation (ii) β Equation (i)

So

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

Also,

$$\begin{array}{l}2.25=\frac{225}{100}=\frac{9}{4};\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\log }_{2.25}}\left( 0.\overline{4} \right)={{\log }_{\left( \frac{9}{4} \right)}}\left( \frac{4}{9} \right)=-1\end{array}$$

Problem26:Β  Find the value of

$$\begin{array}{l}{{2}^{{{\log }_{6}}18}}{{.3}^{{{\log }_{6}}3}}\end{array}$$

Solution:

We can solve above problem by using

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

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

=

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

Problem 27:Β  Find the value of,

$$\begin{array}{l}{{\log }_{\sec \,\,\alpha }}\left( {{\cos }^{3}}\alpha \right)\ \text{where}\ \alpha \in \left( 0,\pi /2 \right)\end{array}$$

Solution: Consider

$$\begin{array}{l}{{\log }_{\sec \,\,\alpha }}\left( {{\cos }^{3}}\alpha \right)=x.\end{array}$$

Therefore, by using formula

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

Let

$$\begin{array}{l}{{\log }_{\sec \,\,\alpha }}{{\cos }^{3}}\alpha =x\end{array}$$

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

Problem 28:Β  If k β N,Β such that

$$\begin{array}{l}{{\log }_{2}}x+{{\log }_{4}}x+{{\log }_{8}}x={{\log }_{k}}x\,and\,\,\forall x\in R’\,\,If\,\,k={{\left( a \right)}^{1/b}}\end{array}$$
then find the value of a + b; a β N, b β N and b is a prime number.

Solution:

By using

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

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

$$\begin{array}{l}k={{\left( a \right)}^{1/b}}\end{array}$$
, we can obtain the value of a + b.

Given,

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

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

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

Β

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

a = 64, b = 11

a + b = 64 + 11 = 75

Problem 29:

$$\begin{array}{l}{{\log}_{e}}\,[{{(1+x)}^{1+x}}{{(1-x)}^{1-x}}]\,=\end{array}$$

Solution:

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

Problem 30: In the expansion of

$$\begin{array}{l}2{{\log}_{e}}x-{{\log}_{e}}(x+1)-{{\log}_{e}}(x-1)\end{array}$$
, the coefficient of x-4 is

Solution:

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

Problem 31:

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

Solution:

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

Problem 32: The coefficient of xn in the expansion of

$$\begin{array}{l}{{\log}_{e}}(1+3x+2{{x}^{2}})\end{array}$$
is

Solution:

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

Problem 33:

$$\begin{array}{l}1+\left(\frac{1}{2}+\frac{1}{3} \right)\,\frac{1}{4}+\left(\frac{1}{4}+\frac{1}{5} \right)\,\frac{1}{{{4}^{2}}}+\left(\frac{1}{6}+\frac{1}{7} \right)\,\frac{1}{{{4}^{3}}}+….\infty =\end{array}$$

Solution:

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

Problem 34:

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

Problem 35:

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

Problem 36:

$$\begin{array}{l}\log \frac{(1+x)^{(1-x) / 2}}{(1-x)^{(1+x) / 2}} \text {is equal to}\\ (1) x+\frac{5 x^{3}}{2.3}+\frac{9 x^{5}}{4.5}+\frac{13 x^{7}}{6.7}+\ldots \ldots . .+\infty\\ (2) x-\frac{5 x^{3}}{2.3}+\frac{9 x^{5}}{4.5}-\frac{13 x^{7}}{6.7}+\ldots \ldots . .+\infty\\ (3) x-\frac{5 x^{3}}{2.3}-\frac{9 x^{5}}{4.5}-\frac{13 x^{7}}{6.7}-\ldots \ldots . .-\infty\\ (4)\ \text{None of these}\\ \text{Solution}: \log \frac{(1+x)^{(1-x) / 2}}{(1-x)^{(1+x) / 2}}\\ \begin{array}{l} =\frac{1}{2}(1-x) \log (1+x)-\frac{1}{2}(1+x) \log (1-x) \\ =\frac{1}{2}[\log (1+x)-\log (1-x)]-\frac{1}{2}[\log (1+x)+\log (1-x)] \\ =\frac{1}{2} \cdot 2\left[\left[x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots \ldots .\right]-\frac{1}{2} \cdot x(-2)\left[\frac{1}{2} x^{2}+\frac{x^{4}}{4}+\ldots . .\right]\right. \\ =x+\left(\frac{1}{3}+\frac{1}{2}\right) x^{2}+\left(\frac{1}{5}+\frac{1}{4}\right) x^{5}+\left(\frac{1}{7}+\frac{1}{6}\right) x^{7}+\ldots \ldots . \\ =x+\frac{5 x^{3}}{2.3}+\frac{9 x^{5}}{4.5}+\frac{13 x^{7}}{6.7}+\ldots \ldots \ldots . \end{array}\\ Answer:[1]\end{array}$$

## Logarithms – Video Lesson 2

Q1

### Who invented logarithms?

John Napier invented logarithms.

Q2

### Give the product rule of logarithms.

Product rule: loga(mn) = logam + logan.

Q3

### Give the quotient rule of logarithms.

Quotient rule: loga(m/n) = logam β logan.

Q4

### Give two applications of logarithms.

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

Q5

### State the power rule of the logarithm.

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