Logarithmics

**Stuart** · March 8th 2006, 05:01 PM

How would I solve:

2log n - log 2n = 2
3 3

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

**ThePerfectHacker** · March 8th 2006, 06:30 PM

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.

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