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

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Answer 1 · 2016-08-04T17:37:11

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

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

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

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

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

$n=-7$ or $n=-2$ 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(n^2+12)=ln(-9n-2)?