1) Gathering the data
[tex]\begin{gathered} P(t)=\text{ 13,158 }\times10^{0.041t} \\ t_0=Current\text{ year} \\ \end{gathered}[/tex]2) To find how many years from the current year will it take, let's plug into that equation:
[tex]\begin{gathered} P(t)=\text{ 13,158 }\times e^{0.041t} \\ 55,000=13158\times e^{0.041t} \\ \frac{55000}{13158}=\frac{13518}{13518}\times e^{0.041t}^{} \\ 4.1799=e^{0.041t}^{} \\ \ln (4.1799)\text{ =}\ln (e^{0.041t}) \\ 1.4302=0.041t \\ t=34.88\cong35 \end{gathered}[/tex]3) So approximately 35 years from the current year.