We have the equations of 3 functions and we want to find which one of them will decay the fastest to 4 grams.
This can be done by setting each expression equal to 4 and find which of them gives the smallest value of t.
For Amount A:
[tex]4 =8e^{-0.1386t} \\ \\
\frac{1}{2}=e^{-0.1386t} \\ \\
ln(\frac{1}{2})=ln(e^{-0.1386t}) \\ \\
-0.693=-0.1386t \\ \\
t=3.775[/tex]
Likewise for Amount B:
[tex]4=10e^{-0.2t} \\ \\
\frac{2}{5}= e^{-0.2t} \\ \\
ln(\frac{2}{5})=ln(e^{-0.2t}) \\ \\
-0.916=-0.2t \\ \\
t=4.58
[/tex]
Similarly for Amount C:
[tex]4=9e^{-0.3t} \\ \\
ln( \frac{4}{9})=ln(e^{-0.3t} ) \\ \\
-0.811=-0.3t \\ \\
t=2.073[/tex]
Thus Amount C will take the minimum time to decay to 4 grams.
