Logarithm

Definition: Logarithm is defined as the inverse function of exponentiation, i.e., $\forall\; a>0$, if:

$\qquad\quad x = a^y$
$\Rightarrow\quad\;\; log_a{x} = y$

Here, clearly $x>0$

Multiplication of log

Theorem: $log_{a_1}{(a_0)}\;log_{a_2}{(a_1)}\cdots log_{a_0}{(a_n)} = 1$

Prerequisites:
Change of base property of log (proof)

Proof:

$\qquad\quad LHS = log_{a_1}{(a_0)}\;log_{a_2}{(a_1)}\cdots log_{a_0}{(a_n)}$

Using change of base property, each term can be converted to base $a_0$:

$\therefore\quad\;\;\text{LHS} = \dfrac{log_{a_0}{(a_0)}}{log_{a_0}{(a_1)}}\dfrac{log_{a_0}{(a_1)}}{log_{a_0}{(a_2)}}\cdots \dfrac{log_{a_0}{(a_{n-1})}}{log_{a_0}{(a_n)}}\dfrac{log_{a_0}{(a_n)}}{log_{a_0}{(a_0)}}$

Since denominator of each term cancels with the numerator of each successive term,
$\therefore\quad\;\;\text{LHS} = 1$

Hence the result.


Recommended:
Log of product
Log of powers
Change of base property

Log Of Powers

Theorem: $\:\:log{(x^y)} = y\; log{(x)}$

Prerequisites:
Logarithm (definition)
Product of the powers property (proof)

Proof:
Let $x = a^z,\;$ where $a$ is the base of the log. Then by definition of log,

$\qquad\quad log_a{(x)} = z\qquad\qquad\qquad \qquad\;\;\;\cdots (1)$

$\begin{align}\text{Now, }\; log_a{(x^y)} &= log_a{\big(a^z\big)^y}\\
&= log_a{(a^{yz})}\qquad\qquad\quad\text{(by product of the power property)}\\
&= yz\qquad\qquad\qquad\qquad\text{(by definition of log)}\\
&= y\; log_a{(x)}\qquad\qquad\quad\text{(from $(1)$)}\end{align}$

Hence the result.

Recommended:
Log of product
Multiplication of log
Change of base property

Log Of Product

Theorem: $log{(xy)} = log{(x)} + log{(y)}$

Prerequisites:
Logarithm (definition)
Sum of the powers property (proof)

Proof:
Let $x = a^u,\;\; y=a^v$  where '$a$' is the base of the log.

$\begin{align}\therefore\quad\;\; log_a{(xy)} &= log_a{(a^ua^v)}\\
&= log_a{(a^{u+v})}\qquad\qquad\qquad\!\!\text{(by sum of powers property of exponentiation)}\\
&= u+v\qquad\qquad\qquad\qquad\text{(by definition of log)}\end{align}$

But, $\quad log_a{(x)} = log_a{(a^u)} = u$
And, $\!\quad log_a{(y)}= log_a{(a^v)} = v$

$\begin{align}\therefore\quad\;\; log_a{(xy)} &= u+v\\
&= log_a{(x)} + log_a{(y)}\end{align}$

Hence the result.


Recommended:
Log of powers
Multiplication of log
Change of base property

Log: Change Of Base Property

Theorem: $log_a{x} = \dfrac{log_b{x}}{log_b{a}}$

Prerequisites:
Logarithm (definition)
Log of powers property (proof)

Proof:
Let $a = b^z$. Then by definition of log,

$\qquad\quad log_b{a} = z\qquad\qquad\qquad\qquad\qquad\qquad\cdots (1)$

Let $log_a{x} = y$, then again by definition of log:

$\qquad\quad\begin{align} x &= a^y\\
&= \big(b^{y}\big)^z\\
&= b^{yz}\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{(by log of powers property)}\end{align}$

Now, taking log on both sides:

$\qquad\quad\begin{align} log_b{x} &= yz\\
&= log_a{x}\;log_b{a}\qquad\qquad\qquad\qquad\text{(from (1))}\end{align}$

$\Rightarrow\quad\;\; log_a{x} = \dfrac{log_b{x}}{log_b{a}}$

Hence the result.


Recommended:
Log of product
Log of powers
Multiplication of log