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!

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