Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.6 Applications and Models of Exponential Growth and Decay - 4.6 Exercises - Page 480: 21

$16\ days$

Step 1. Given $A(t)=A_0e^{-0.087t}$, decay to $25\%$ means $A(t)=0.25A_0$ Step 2. we have $0.25A_0=A_0e^{-0.087t} \longrightarrow e^{-0.087t}=0.25 \longrightarrow -0.087t=ln(0.25) \longrightarrow t=\frac{1}{-0.087}ln(0.25)\approx16\ days$