Chapter 6 - Section 6.8 - Exponential Growth and Decay Models; Newton's Law: Logistic Growth and Decay Models - 6.8 Assess Your Understanding - Page 487: 21

$t=26.6$ days

$A(t)=A_{0}e^{kt},$ $A_{0}$ is the Initial amount of radioactive material. $k$ is negative number. In this case. $A_{0}=1,$ $ A(8)=0.5, t=8$ days $A(8)=e^{8k}=0.5,$ $A(8)=e^{8k}=0.5,$ $k=\ln{0.5}$ $k=-0.0866,$ Thus, $0.1=e^{-0.0866t},$ $e^{-0.0866t}=0.1,$ $-0.0866t=\ln{0.1},$ $t=26.6$ days