\(f(t)=(A-G)\cdot e^{-k\cdot t}+G |-G\)
\(f(t)-G=(A-G)\cdot e^{-k\cdot t} |:(A-G)\)
\( e^{-k\cdot t}= \frac{f(t)-G}{A-G}\)
\(-k\cdot t\cdot ln(e)=ln(\frac{f(t)-G}{A-G})\) mit \(ln(e)=1\)
\(-k\cdot t=ln(\frac{f(t)-G}{A-G})|:t\)
\(-k=\frac{ln(\frac{f(t)-G}{A-G})}{t}\)
\(k=-\frac{ln(\frac{f(t)-G}{A-G})}{t}\)