How would I solve:
2log n - log 2n = 2
3 3
and the 3's should be under that logs right side.
$\displaystyle 2\log_3 n-\log_3(2n)=2$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.