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

1 Answer
Aug 4, 2016

Answer:

#n=-7# or #n=-2#

Explanation:

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