Question

Knowledge CheckQuestion 28An initial population of 60 fish is introduced into a lake. This fish population grows according to a continuous exponential growth model. There are 144 fishin the lake after 8 years

Answer 1

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

$P=P_0e^{rt}$

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).

$144=60\cdot e^{r\cdot8}$

Now, we solve for r

$\begin{gathered} \frac{144}{60}=e^{8r} \\ e^{8r}=2.4 \end{gathered}$

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

$\begin{gathered} \ln e^{8r}=\ln 2.4 \\ 8r=\ln 2.4 \\ r=\frac{\ln 2.4}{8} \\ r\approx0.1094 \end{gathered}$

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

(a)

$P=60\cdot e^{0.1094t}$

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).

$P=60\cdot e^{0.1094t}$

Where t = 19, which means 19 years.

$\begin{gathered} P=60\cdot e^{0.1094\cdot19} \\ P\approx480 \end{gathered}$

Therefore, there are 480 fish after 19 years. (b)