Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.6 Applications and Models of Exponential Growth and Decay - 4.6 Exercises - Page 480: 20

$0.88\ g$

Step 1. Given $A_0=1\ g, A(2)=0.95\ g$, we have $0.95=e^{-2k} \longrightarrow e^{-2k}=0.95 \longrightarrow -2k=ln(0.95) \longrightarrow k=\frac{1}{-2}ln(0.95)\approx0.0256$ Step 2. After $t=5\ yr$, we have $A(5)=e^{-0.0256(5)}\approx0.88\ g$