How do you solve #log_12(v^2+35)=log_12(-12v-1)#?

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Answer 1 · 2016-07-10T00:19:57

#log_12(v^2 + 35) - log_12(-12v - 1) = 0#

#log_12((v^2 + 35)/(-12v - 1)) = 0#

#(v^2 + 35)/(-12v - 1) = 12^0#

#v^2 + 35 = 1(-12v - 1)#

#v^2 + 35 = -12v - 1#

#v^2 + 12v + 36 = 0#

#(v + 6)^2 = 0#

#(v + 6)(v + 6) = 0#

#v = -6 and -6#

Checking in the original solution, we find this solution works.

This equation has a solution set #{-6}#.

Hopefully this helps!