Answer
$ t\approx 34.2$ years from now.
Work Step by Step
Model: $ A=A_{0}e^{kt}$
Unknown: $ k $, when $ A=1000,\ A_{0}=1400,\ t=5$ years
$1000=1400e^{k\cdot 5}\qquad... /\div 1400$
$0.714286=e^{k\cdot 5}\qquad.../\ln(...)$
$-0.336472=5k\qquad.../\div(5)$
$ k\approx-0.067294$
So, our model is
$ A=1000e^{-0.067294t}$, if we count t=0 as today.
Now we find $ t $ for $ A=100$
$100=1000e^{-0.067294t}\qquad... /\div A_{0}$
$0.1=e^{-0.067294t}\qquad.../\ln(...)$
$\ln 0.1 =-0.067294t\qquad.../\div(-0.067294)$
$\displaystyle \frac{\ln 0.1}{-0.067294}=t $
$ t\approx 34.2$ years from now.
