- Logarithm of a to base b is x if $b^x=a$

Where a and b are positive real numbers and $b \ne 1$ - It is denoted as $log_b a=x$

$log_2 4 =2$ as $2^2=4$

$log_{10} 1000 =3$ as $10^3=1000$

$log_{25} 625 =2$ as $25^2=625$

$f(x)=log_a x$

where a is called the base and x can be any positive real number. The base b should be greater than 0 and not equal to 1.

$f(x)=log_{10} x$

$f(x)=log_e x$

Where e is Euler's number, approximately equal to 2.71828

Graph of $f(x)=log_{e} x$

Graph of $f(x)=log_2 x$

Comparison Graph $f(x)=log_{10}x$ ,$f(x)=log_{e} x$ ,$f(x)=log_2 x$

(1) The domain of log function is R+

(2) The range of log function is the set of all real numbers.

(3) The point (1, 0) is always on the graph of the log function.

(4) The log function is ever increasing, i.e., as we move from left to right the graph rises above.

(5) For x very near to zero, the value of log x can be made lesser than any given real number. In other words in the fourth quadrant the graph approaches y-axis (but never meets it).

Graph of $f(x)=log_{.2}x$

Comparison Graph $f(x)=log_{.5}x$ ,$f(x)=log_{.2}x$, $f(x)=log_{.8}x$

(1) The domain of log function is R+

(2) The range of log function is the set of all real numbers.

(3) The point (1, 0) is always on the graph of the log function.

(4) The log function is ever decreasing, i.e., as we move from left to right the graph goes down

It is of interest to observe that the two curves are the mirror images of each other reflected in the line y = x

$a^{log_a m}=m$

Proof:

From definition of Logarithm

$log_a m =x$

$a^x =m$

substituting values

$a^{log_a m}=m$

(2) $log_a x^n=nlog_a x$

Proof:

$log_a x^n=t$

Using definition of Logarithm

$x^n =a^t$

Now taking 1/n power

$x=a^{t/m}$

Using Definition of Logarithm

$log_a x=t/n$

or $t=nlog_a x$

or $log_a x^n =nlog_a x$

(3) $log_{a^m} x=\frac {1}{m}log_a x$

Proof:

$log_{a^m} x=t$

Using definition of Logarithm

$log_a x =mt$

$t=\frac {1}{m} log_a x $

or $log_{a^m} x =\frac {1}{m} log_a x $

(4) $log_{a} p + log_{a} q=log_a pq$

Proof:

Let $log_{a} p= m$ or $p=a^m$

Let $log_{a} q= n$ or $q=a^n$

Multiplying both

$pq= a^m a^n$

$a^{m+n}= pq$

Using Definition of Logarithm

$m +n = log_a pq$

$log_{a} p + log_{a} q=log_a pq$

(5)$log_{a} p - log_{a} q=log_a \frac {p}{q}$

Proof:

Let $log_{a} p= m$ or $p=a^m$

Let $log_{a} q= n$ or $q=a^n$

Dividing both

$\frac {p}{q}= \frac {a^m}{a^n}$

$a^{m-n}= \frac {p}{q}$

Using Definition of Logarithm

$m - n = log_a \frac {p}{q}$

$log_{a} p - log_{a} q=log_a \frac {p}{q}$

(6)$log_{a} c = \frac {log_{b} c}{log_{b} a}$

Proof:

Let $log_{a} c=x$ or $a^x=c$ --(1)

Let $log_{b} c=y$ or $b^y=c$ --(2)

Let $log_{b} a=z$ or $b^z=a$ --(3)

Substituting the value of a from (3) in (1)

$b^{xz}=c$

Substituting the value of c from (2)

$b^{xz}=b^y$

So

xz=y

or x= y/z

or

$log_{a} c = \frac {log_{b} c}{log_{b} a}$

- Logarithmic differentiation is a technique used to differentiate functions by taking the natural logarithm (ln) of both sides of an equation and then using the properties of logarithms to simplify the differentiation process.
- This method is particularly useful for functions where direct differentiation is complex.

Find the derivative of \( y = x^x \).

We can solve using Logarithmic differentiation

Take the natural logarithm of both sides:

\( \ln(y) = \ln(x^x) \).

\( \ln(y) = x \ln(x) \).

Differentiate both sides with respect to \( x \)

\[ \frac{1}{y} \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} \] \[ \frac{dy}{dx} = y \cdot (\ln(x) + 1) \] Substituting the value of y

\[ \frac{dy}{dx} = x^x (\ln(x) + 1) \]

Find the derivative of \( y = x^2 \cdot e^x \cdot \ln(x) \).

Derivative can be found in many ways. Let use the technique of Logarithmic differentiation

Take the natural logarithm of both sides:

\( \ln(y) = \ln(x^2 \cdot e^x \cdot \ln(x)) \).

\( \ln(y) = \ln(x^2) + \ln(e^x) + \ln(\ln(x)) \).

\( \ln(y) = 2\ln(x) + x + \ln(\ln(x)) \).

Differentiate both sides with respect to \( x \):

\[ \frac{1}{y} \frac{dy}{dx} = \frac{2}{x} + 1 + \frac{1}{x \ln(x)} \] \[ \frac{dy}{dx} = y \left( \frac{2}{x} + 1 + \frac{1}{x \ln(x)} \right) \] Substituting the value of y

\[ \frac{dy}{dx} = x^2 e^x \ln(x) \left( \frac{2}{x} + 1 + \frac{1}{x \ln(x)} \right) \]

Find the derivative of \( y = \frac{x^3}{\ln(x)} \).

Derivative can be found in many ways. Let use the technique of Logarithmic differentiation

Take the natural logarithm of both sides:

\( \ln(y) = \ln\left(\frac{x^3}{\ln(x)}\right) \).

\( \ln(y) = \ln(x^3) - \ln(\ln(x)) \).

\( \ln(y) = 3\ln(x) - \ln(\ln(x)) \).

Differentiate both sides with respect to \( x \):

\[ \frac{1}{y} \frac{dy}{dx} = \frac{3}{x} - \frac{1}{x \ln(x)} \] \[ \frac{dy}{dx} = y \left( \frac{3}{x} - \frac{1}{x \ln(x)} \right) \] Substituting the value of y

\[ \frac{dy}{dx} = \frac{x^3}{\ln(x)} \left( \frac{3}{x} - \frac{1}{x \ln(x)} \right) \]

**Notes**-
**NCERT Solutions & Assignments**- NCERT Solution for Continuity And differentiability Exercise 5.1
- NCERT Solution for Continuity And differentiability Exercise 5.2
- NCERT Solution for Continuity And differentiability Exercise 5.3
- NCERT Solution for Continuity And differentiability Exercise 5.4
- Continuity and Differentiability class 12 Important questions

Class 12 Maths Class 12 Physics Class 12 Chemistry Class 12 Biology