Question

log 12 ( v 2 + 35) = log 12 ( −12v − 1)

Answer 1

\[\huge \log_{12} (v^2 +35) = \log_{12} (-12v-1)\]
you can simplify this to:\[\huge v^2 + 35 = -12v -1\]
\[\huge v^2 +12v + 36 = 0\]
\[\huge (x+6)(x+6) = 0 , x = -6\]