Given data:
![\begin{gathered} P_0=4000 \\ t=12 \\ P=12500 \\ \\ P=P_0e^(kt) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/s4afkqn4u4oyfknkr0vy7qgfxdb6r6xh7m.png)
Use the given data in the model and solve k (growth rate):
![12500=4000e^(12k)](https://img.qammunity.org/2023/formulas/mathematics/college/k3xt6q0c1uv72qwwzdaw9awezauekvsul6.png)
Divide both sides of the equation into 4000:
![\begin{gathered} (12500)/(4000)=(4000)/(4000)e^(12k) \\ \\ (25)/(8)=e^(12k) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tpeijw9l41qav6lpir9dk476edymidcusc.png)
Find the natural logarithm of both sides of the equation:
![\begin{gathered} \ln ((25)/(8))=\ln (e^(12k)) \\ \\ \ln ((25)/(8))=12k \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ytf2f304o0ptbcwb49okdnxvup5lld173v.png)
Divide both sides of the equation by 12:
![\begin{gathered} (\ln ((25)/(8)))/(12)=(12)/(12)k \\ \\ (\ln((25)/(8)))/(12)=k \\ \\ \\ k=(\ln((25)/(8)))/(12) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/791rogv3femwb12be8fc2n2hdmx7e9gvh0.png)
Evaluate:
![k\approx0.095](https://img.qammunity.org/2023/formulas/mathematics/college/qsw5fj26r50tt2foigoieqzestmyfvf7oy.png)
Then, the growth rate is 0.095 (9.5%)