Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.6 Applications and Models of Exponential Growth and Decay - 4.6 Exercises - Page 480: 19

$3.6\ g$

Step 1. Given $A_0=12\ g, A(4)=6.0\ g$, we have $6=12e^{-4k} \longrightarrow e^{-4k}=\frac{1}{2} \longrightarrow -4k=ln(\frac{1}{2}) \longrightarrow k=\frac{1}{-4}ln(\frac{1}{2})\approx0.17329$ Step 2. After $t=7\ yr$, we have $A(7)=12e^{-0.17329(7)}\approx3.6\ g$