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