Answer
(a) $440\ g$
(b) $387\ g$
(c) $264\ g$
(d) $22\ yr $
Work Step by Step
Given $A(t)=500e^{-0.032t}$, we have:
(a) For $t=4\ yr$, $A(4)=500e^{-0.032(4)}\approx440\ g$
(b) For $t=8\ yr$, $A(8)=500e^{-0.032(8)}\approx387\ g$
(c) For $t=20\ yr$, $A(20)=500e^{-0.032(20)}\approx264\ g$
(d) For half-life, let $A(t)=500/2=250$, we have $250=500e^{-0.032t} \longrightarrow e^{-0.032t}=\frac{1}{2} \longrightarrow -0.032t=ln(\frac{1}{2}) \longrightarrow t=\frac{1}{-0.032}ln(\frac{1}{2})\approx22\ yr $
