Answer: 129.33 g
Step-by-step explanation:
[tex]$Let $p=A \cdot e^{n t}$ \\$(p=$ present amount, $A=$ initial amount, $n=$ decay rate, $t=$ time)[/tex]
[tex]\begin{aligned}&\Rightarrow \text { Given } p=\frac{A}{2} \ a t \ t=1590 \\&\Rightarrow \frac{1}{2}=e^{1590n} \Rightarrow n=\frac{\ln (1 / 2)}{1590}=-0.00043594162\end{aligned}[/tex]
[tex]$If $A=200 \mathrm{mg}$ and $t=1000$ then,$\begin{aligned}P &=200 e^{\left(\frac{\ln \left(1/2\right)}{1590}\right) \cdot 1000} \text \\\\&=129.33 \text { grams}\end{aligned}$[/tex]
