Answer
about 7 years
Work Step by Step
The decay model: $A=A_{0}e^{kt}$ ($ k < 0 )$
We find k from the half life information
$0.5A_{0}=A_{0}e^{k\cdot 22}\qquad/\div A_{0}$
$0.5=e^{22k}\qquad $ ... apply ln( ) to both sides
$\ln 0.5=22k$
$ k=\displaystyle \frac{\ln 0.5}{22}\approx$-0.0315066900255$\approx-0.0315$
The decay model is : $A=A_{0}e^{-0.0315t}$
We want t (in years) when $A=0.80A_{0}$
$0.8A_{0}=A_{0}e^{-0.0315t}\qquad/\div A_{0}$
$0.8=e^{-0.0315t}\qquad $ ... apply ln( ) to both sides
$\ln 0.8=-0.0315t\qquad \div(-0.0315)$
$ t=\displaystyle \frac{\ln 0.8}{-0.0315}\approx$7.08392226394$\approx$ 7 years