# How do you solve ln(-3+3n)=ln(n^2-n)?

##### 1 Answer
Aug 10, 2016

$n = 3$

#### Explanation:

${\log}_{e} \left(- 3 + 3 n\right) - {\log}_{e} \left({n}^{2} - n\right) = \log \left(n - 1\right) + {\log}_{e} 3 - {\log}_{e} \left(n - 1\right) - {\log}_{e} n = 0$

or

${\log}_{e} 3 - {\log}_{e} n = {\log}_{e} 1$

Finally

${\log}_{e} \left(\frac{3}{n}\right) = {\log}_{e} 1$

or

$\frac{n}{3} = 1 \to n = 3$ which is a feasible solution