How do you solve $ln (- 3 + 3 n) = ln (n^{2} - n)$ $ln(-3+3n)=ln(n^2-n)$ ?

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How do you solve $ln (- 3 + 3 n) = ln (n^{2} - n)$ $ln(-3+3n)=ln(n^2-n)$ ?

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Answer 1 · 2016-08-10T08:31:26

$n = 3$ $n = 3$

Explanation:

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

or

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

Finally

${log}_{e} (\frac{3}{n}) = {log}_{e} 1$ $log_e(3/n) = log_e 1$

or

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

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How do you solve $ln (- 3 + 3 n) = ln (n^{2} - n)$ $ln(-3+3n)=ln(n^2-n)$ ?

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