How would I solve:

2log n - log 2n = 2

3 3

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

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- Mar 8th 2006, 05:01 PMStuartLogarithmics
How would I solve:

2log n - log 2n = 2

3 3

and the 3's should be under that logs right side. - Mar 8th 2006, 06:30 PMThePerfectHackerQuote:

Originally Posted by**Stuart**

Thus, since $\displaystyle 2\log_3 n=\log_3 n^2$

$\displaystyle \log_3 n^2-\log_3 (2n)=2$

The quotient rule,

$\displaystyle \log_3\left(\frac{n^2}{2n}\right)=2$

Cancel out the $\displaystyle n$'s,

$\displaystyle \log_3\left(\frac{n}{2}\right)=2$

By the definition of logs,

$\displaystyle \frac{n}{2}=3^2$

Thus,

$\displaystyle n=18$

Q.E.D.