Answer
$\approx250.39$
Work Step by Step
We know that $34=50e^{k\cdot30}\\0.68=e^{30k}\\30k=\ln{0.68}\\k=\frac{\ln{0.68}}{30}\approx-0.0128554$
Thus $2=50e^{-0.0128554\cdot t}\\0.04=e^{-0.0128554\cdot t}\\-0.0128554=\ln0.04\\t=\frac{\ln{0.04}}{-0.0128554}\approx250.39$
