Answer
$ t\approx 7.1$ years
Work Step by Step
Model: $ A=A_{0}e^{kt}$
Unknown: $ k $, when $ A=0.5A_{0},\ t=22$
$0.5A_{0}=A_{0}e^{k\cdot 22}\qquad... /\div A_{0}$
$0.5=e^{k\cdot 22}\qquad.../\ln(...)$
$-0.693147=22k\qquad.../\div(22)$
$ k\approx-0.031507$
So, our model is
$ A=A_{0}e^{-0.031507t}$
Now we find $ t $ for $ A=0.8A_{0}$
$0.8A_{0}=A_{0}e^{-0.031507t}\qquad... /\div A_{0}$
$0.8=e^{-0.031507t}\qquad.../\ln(...)$
$\ln 0.8 =-0.031507t\qquad.../\div(-0.031507)$
$\displaystyle \frac{\ln 0.8}{-0.031507}=t $
$ t\approx 7.1$ years
