$
\eqalign{
& stap\,\,1. \cr
& 0 = a \cdot {}^5\log (6 - c) \cr
& a = 0\,\,(v.n.)\,\,of\,\,{}^5\log (6 - c) = 0 \cr
& 6 - c = 1 \cr
& c = 5 \cr
& stap\,\,2. \cr
& - 12 = a \cdot {}^5\log (130 - c) \cr
& - 12 = a \cdot {}^5\log (130 - 5) \cr
& - 12 = a \cdot {}^5\log (125) \cr
& - 12 = a \cdot 3 \cr
& a = - 4 \cr}
$
Opgelost!
WvR
12-10-2019