To find the answer, we will need to replace the population P by 50,000 and solve the initial equation for t because t is the number of years after 2012.
So, we get:
[tex]50,000=25,000e^{0.03t}[/tex]
Now, we need to remember some properties of the logarithms:
[tex]\ln e^a=a[/tex]
Then, we can solve for t as:
[tex]\begin{gathered} 50,000=25,000e^{0.03t} \\ \frac{50,000}{25,000}=\frac{25,000e^{0.03t}}{25,000} \\ \\ 2=e^{0.03t} \end{gathered}[/tex]
So, using the property, we get:
[tex]\begin{gathered} \ln 2=\ln e^{0.03t} \\ \ln 2=0.03t \end{gathered}[/tex]
Finally, dividing by 0.03 into both sides, we get that the number of years after 2012 that the population will be 50,000 is:
[tex]\begin{gathered} \frac{\ln 2}{0.03}=\frac{0.03t}{0.03} \\ \frac{\ln 2}{0.03}=t \end{gathered}[/tex]
Answer: t = ln2/0.03
