We know that
• The initial population is 60.
,
• There are 144 fish after 8 years.
To solve this problem, we have to use the population growth exponential expression
![P=P_0e^{rt}](https://tex.z-dn.net/?f=P%3DP_0e%5E%7Brt%7D)
Where P0 is the initial population, r is the rate of growth, t is time in years, and P is the final population.
Let's use the given information to find the rate of growth (r).
![144=60\cdot e^{r\cdot8}](https://tex.z-dn.net/?f=144%3D60%5Ccdot%20e%5E%7Br%5Ccdot8%7D)
Now, we solve for r
![\begin{gathered} \frac{144}{60}=e^{8r} \\ e^{8r}=2.4 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7B144%7D%7B60%7D%3De%5E%7B8r%7D%20%5C%5C%20e%5E%7B8r%7D%3D2.4%20%5Cend%7Bgathered%7D)
In order to solve for r, we have to apply a natural logarithm on each side so we can eliminate the power that contains r
![\begin{gathered} \ln e^{8r}=\ln 2.4 \\ 8r=\ln 2.4 \\ r=\frac{\ln 2.4}{8} \\ r\approx0.1094 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cln%20e%5E%7B8r%7D%3D%5Cln%202.4%20%5C%5C%208r%3D%5Cln%202.4%20%5C%5C%20r%3D%5Cfrac%7B%5Cln%202.4%7D%7B8%7D%20%5C%5C%20r%5Capprox0.1094%20%5Cend%7Bgathered%7D)
Note that we use four decimal digits, that's because we'll get more precision.
Once, we have the rate of growth we can write the exponential function that represents the situation
<h2>(a)</h2>
![P=60\cdot e^{0.1094t}](https://tex.z-dn.net/?f=P%3D60%5Ccdot%20e%5E%7B0.1094t%7D)
On the other hand, to find the number of fish there are after 19 years, we have to use the exponential expression we found in (a).
![P=60\cdot e^{0.1094t}](https://tex.z-dn.net/?f=P%3D60%5Ccdot%20e%5E%7B0.1094t%7D)
Where t = 19, which means 19 years.
![\begin{gathered} P=60\cdot e^{0.1094\cdot19} \\ P\approx480 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%3D60%5Ccdot%20e%5E%7B0.1094%5Ccdot19%7D%20%5C%5C%20P%5Capprox480%20%5Cend%7Bgathered%7D)
<h2>Therefore, there are 480 fish after 19 years. (b)</h2>