Answer
(a) $404\ g$
(b) $327\ g$
(c) $173\ g$
(d) $13\ yr $
Work Step by Step
Given $A(t)=500e^{-0.053t}$, we have:
(a) For $t=4\ yr$, $A(4)=500e^{-0.053(4)}\approx404\ g$
(b) For $t=8\ yr$, $A(8)=500e^{-0.053(8)}\approx327\ g$
(c) For $t=20\ yr$, $A(20)=500e^{-0.053(20)}\approx173\ g$
(d) For half-life, let $A(t)=500/2=250$, we have $250=500e^{-0.053t} \longrightarrow e^{-0.053t}=\frac{1}{2} \longrightarrow -0.053t=ln(\frac{1}{2}) \longrightarrow t=\frac{1}{-0.053}ln(\frac{1}{2})\approx13\ yr $
