we have the equation
[tex]y=y_0e^{-0.0415t}[/tex]
Half-time is the time when y=y0/2
substitute
[tex]\begin{gathered} \frac{y_0}{2}=y_0e^{-0.0415t} \\ \\ simplify \\ \frac{1}{2}=e^{-0.0415t} \end{gathered}[/tex]
Solve for t
Apply ln on both sides
[tex]\begin{gathered} ln(\frac{1}{2})=ln(e^{-0.0415t}) \\ \\ ln(\frac{1}{2})=-0.0415t \\ t=\frac{ln(\frac{1}{2})}{-0.0415} \\ \\ t=16.7\text{ days} \\ \end{gathered}[/tex]
