[tex]\begin{gathered} P=25000e^{0.03t} \\ P=50000 \\ 25000e^{0.03t}=50000 \end{gathered}[/tex][tex]\begin{gathered} \text{divide both side by 25000} \\ e^{0.03t}=\frac{50000}{25000}\Rightarrow e^{0.03t}=2 \\ \text{Take the N}aturalLogarithm\text{ of both side: } \\ \ln e^{0.03t}=\ln 2 \\ 0.03t=\ln 2 \\ t=\frac{\ln 2}{0.03} \end{gathered}[/tex]
Hence option A is the correct answer
