Answer
$A(72)=1.703,$
and he must shock again $t=171.912$ hours later
Work Step by Step
$A(t)=A_{0}e^{kt},$
$A_{0}$ is the Initial amount of radioactive material.
$k$ is negative number.
In this case. $A_{0}=2.5,$ $ t=24$ hours
$A(24)=2.5e^{24k}=2.2,$
$A(24)=e^{24k}=0.88,$
$24k=\ln{0.88}$
$k=-0.00533,$
Thus,
$A(72)=2.5e^{72\times -0.00533},$
$A(72)=1.703,$
$1=2.5e^{-0.00533t},$
$e^{-0.00533t}=0.4,$
$-0.00533t=\ln{0.4},$
$t=171.912$ hours later
