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