Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.6 Applications and Models of Exponential Growth and Decay - 4.6 Exercises - Page 480: 18

$12\ yr $

For half-life, let $A(t)=\frac{1}{2}A_0$, we have $\frac{1}{2}A_0=A_0e^{-0.056t} \longrightarrow e^{-0.056t}=\frac{1}{2} \longrightarrow -0.056t=ln(\frac{1}{2}) \longrightarrow t=\frac{1}{-0.056}ln(\frac{1}{2})\approx12\ yr $