Processing math: 100%

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

No comments:

Post a Comment