Answer
$A(24)=7.35,$
$t=76.71$ hour
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. $A_{0}=25,$ $ A(10)=15, t=10$ hours
$A(10)=25e^{10k}=15,$
$A(10)=e^{10k}=0.6,$
$10k=\ln{0.6}$
$k=-0.051,$
Thus,
$A(24)=25e^{24\times -0.051},$
$A(24)=7.35,$
$0.5=25e^{-0.051t},$
$e^{-0.051t}=0.02,$
$-0.051t=\ln{0.02},$
$t=76.71$ hour
