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