\(x= x_{0} e^{kt} ,where~ x ~is~ the ~ |amount\ of \material\ at\ any~~time~t\)
\(and~ x_{0} ~the~original~amount.\)
\(when~x= \frac{12.5}{100} x_{0} }= \frac{125}{1000} x_{0}\)
\(\frac{x}{ x_{0} } = \frac{125}{1000}= \frac{1}{8}\)
\(when t=19.8 hrs,
x= x_{0} e^{19.8 k},
\frac{x}{ x_{0} } = e^{19.8 k},
\frac{1}{8} = e^{19.8k} ,
ln \frac{1}{8} =19.8 k,\)
\(ln1-ln8=19.8k,
0-ln 2^{3} =19.8 k,
-3 ln2=19.8k
k= \frac{-3 ln2}{19.8} = \frac{- ln 2}{6.6}\)\(x= x_{0} e^{ \frac{-ln2}{6.6}t } ,
\frac{x}{ x_{0} } = e^{ \frac{- ln 2}{6.6}t }\)\(\frac{1}{2} = e^{ \frac{- ln2}{6.6}t } = \frac{1}{ e^{ \frac{ln 2}{6.6} t} } ,
2= e^{ \frac{ln2}{6.6}t } ,
ln2=ln e^{ \frac{ln 2}{6.6}t } = \frac{ln 2}{6.6} t~ln e= \frac{ln2}{6.6} t,\)
