Chapter 3 - Section 3.5 - Exponential Growth and Decay; Modeling Data - Exercise Set - Page 506: 21

$12.6$ years

Model: $ A=A_{0}e^{kt}$ Unknown: $ t $, when $ A=0.5A_{0},\ k=-0.055$ $0.5A_{0}=A_{0}e^{-0.055t}\qquad... /\div A_{0}$ $0.5=e^{-0.055t}\qquad.../\ln(...)$ $-0.693147=-0.055t\qquad.../\div(-0.055)$ $ t\approx 12.6$ years $12.6$ years