## Log Base 2

What is the value of
$\dfrac{log_2 3\cdot log_4 5\cdot log_6 7}{log_4 3\cdot log_6 5\cdot log_8 7}$
Source: NCTM Mathematics Teacher, November 2006

SOLUTION
We change all the different bases to base $10$
$log_2 3=log\,3/log\,2$
$log_4 5=log\,5/log\,4$
$log_6 7=log\,7/log\,6$
$log_4 3=log\,3/log\,4$
$log_6 5=log\,5/log\,6$
$log_8 7=log\,7/log\,8$

$\dfrac{log_2 3\cdot log_4 5\cdot log_6 7}{log_4 3\cdot log_6 5\cdot log_8 7}=\dfrac{log\,3/log\,2\cdot log\,5/log\,4\cdot log\,7/log\,6}{log\,3/log\,4\cdot log\,5/log\,6\cdot log\,7/log\,8}$
$=\dfrac{1/log\,2}{1/log\,8}$
$=log\,8/log\,2$
$=log\,2^3/log\,2$
$=3\,log\,2/log\,2$
$=3$

Answer: $3$