How do you solve $ln (n^{2} + 12) = ln (- 9 n - 2)$ $ln(n^2+12)=ln(-9n-2)$ ?

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Answer 1 · 2016-08-04T17:37:11

$n = - 7$ $n=-7$ or $n = - 2$ $n=-2$

raising the equation to $e$ $e$ we can drop the $log$ $log$ so

$n^{2} + 12 = - 9 n - 2$ $n^2+12 = -9n-2$

$n^{2} + 9 n - 14 = 0$ $n^2+9n-14=0$

$(n + 2) (n + 7) = 0$ $(n+2)(n+7)=0$

$n = - 7$ $n=-7$ or $n = - 2$ $n=-2$ are potential candidates.

#now you plug in and make sure
and it turns out that they both work

How do you solve ln(n^2+12)=ln(-9n-2)ln(n2+12)=ln(−9n−2)ln(n^2+12)=ln(-9n-2)?