b.
To determine the half life we plug half the original amount in the expression given and solve for t:
[tex]\begin{gathered} \frac{45}{2}=45e^{-0.0045t} \\ e^{-0.0045t}=\frac{45}{2\cdot45} \\ e^{-0.0045t}=\frac{1}{2} \\ \ln e^{-0.0045t}=\ln (\frac{1}{2}) \\ -0.0045t=\ln (\frac{1}{2}) \\ t=\frac{1}{-0.0045}\ln (\frac{1}{2}) \\ t=154.03 \end{gathered}[/tex]Therefore the half-life of the substance is approximately 154 years.
c.
To determine how much of the substance will be after 2500 years we just plug this value in the expression:
[tex]\begin{gathered} A=45e^{-0.0045\cdot2500} \\ A=5.85\times10^{-4} \end{gathered}[/tex]Therefore after 2500 years there will be:
[tex]5.85\times10^{-4}\text{ gr}[/tex]of the substance.
