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: 19

$A(120)=0.263,$ $t=396.1$ minutes

$A(t)=A_{0}e^{kt},$ $A_{0}$ is the Initial amount. $k$ is negative number. In this case. $A_{0}=0.4,$ $ A(30)=0.36, t=30$ minutes $A(30)=0.4e^{30k}=0.36,$ $A(30)=e^{30k}=0.9,$ $30k=\ln{0.9}$ $k=-0.0035,$ Thus, for $t=120,$ minutes $A(120)=0.4e^{120\times -0.0035},$ $A(120)=0.263,$ $0.1=0.4e^{-0.0035t},$ $e^{-0.0-35t}=0.25,$ $-0.0035t=\ln{0.25},$ $t=396.1$ minutes