continuouslyUsing the formula for a compounded continously
[tex]P=P_0\cdot e^{r\cdot t}[/tex]where P is the amount on the account after t years compounded at an interest rate r when Po is invested in an account.
then,
[tex]\begin{gathered} 8500=P_0\cdot e^{0.095\cdot14} \\ 8500=P_{0^{}}\cdot e^{1.33} \\ P_0=\frac{8500}{e^{1.33}} \\ P_0=2248.056\approx2248.06 \end{gathered}[/tex]