Answer
$250.39$ days.
Work Step by Step
1. Set up the model $A(t)=A_0e^{-kt}$ with $A_0=50mg$.
2. With $t=30\ days, A=34mg$, we have $34=50e^{-30k}$, thus $k=-\frac{ln(34/50)}{30}\approx0.0128554$ and $A(t)=50e^{-0.0128554t}$.
3. For $A=2mg$, we have $2=50e^{-0.0128554t}$ and $t=\frac{ln(2/50)}{-0.0128554}\approx250.39$ days.
