Answer
$ t\approx 17,121.7$ years
Work Step by Step
Model: $ A=A_{0}e^{kt}$
Unknown: $ k $, when $ A=0.5A_{0},\ t=7340$
$0.5A_{0}=A_{0}e^{k\cdot 7340}\qquad... /\div A_{0}$
$0.5=e^{k\cdot 7340}\qquad.../\ln(...)$
$-0.693147=7340k\qquad.../\div(7340)$
$ k\approx-0.000094$
So, our model is
$ A=A_{0}e^{-0.000094t}$
Now we find $ t $ for $ A=0.2A_{0}$
$0.2A_{0}=A_{0}e^{-0.000094t}\qquad... /\div A_{0}$
$0.2=e^{-0.000094t}\qquad.../\ln(...)$
$\ln 0.2 =-0.000094t\qquad.../\div-0.000094$
$\displaystyle \frac{\ln 0.2}{-0.000094}=t $
$ t\approx 17,121.7$ years
