# Logarithmics

• March 8th 2006, 05:01 PM
Stuart
Logarithmics
How would I solve:

2log n - log 2n = 2
3 3

and the 3's should be under that logs right side.
• March 8th 2006, 06:30 PM
ThePerfectHacker
Quote:

Originally Posted by Stuart
How would I solve:

2log n - log 2n = 2
3 3

and the 3's should be under that logs right side.

$2\log_3 n-\log_3(2n)=2$
Thus, since $2\log_3 n=\log_3 n^2$
$\log_3 n^2-\log_3 (2n)=2$
The quotient rule,
$\log_3\left(\frac{n^2}{2n}\right)=2$
Cancel out the $n$'s,
$\log_3\left(\frac{n}{2}\right)=2$
By the definition of logs,
$\frac{n}{2}=3^2$
Thus,
$n=18$
Q.E.D.