Chapter 9 - Section 9.6 - Exponential Growth and Decay; Modeling Data - Exercise Set - Page 741: 28

$17043.7$ years

$A=A_{0}e^{kt}$ For $A=0.5, A_{0}=1, t=7340, $ year $0.5=1e^{7340k} $ $\ln0.5=7340k $ $k=-0.00009443$ decay rate or $0.009443\%$ decay per year. Thus, $0.2=e^{-0.00009443t}$ $\ln 0.2 =-0.00009443t$ $t=17043.7$ years