Question

A rare species of rabbit lives on an island. Their population decreases exponentially according to the model N(t)=a×e^−kt, where t is time measured in years, N(t) is the population at time t, and a, k are constants. Five years ago, 232 rabbits were living on the island, while now they are 146. After how many more years (to the nearest year) will there be only two rabbits left on the island?

Answer 1

To find after how many more years there will be only two rabbits left, we use the exponential decay model N(t)=a×e^(-kt). First, we find the constants a and k using the known population sizes at two different times. Then, we solve for the time t when N(t) will be 2 and round to the nearest year.

Step-by-step explanation:

To determine after how many more years there will be only two rare rabbits left on the island, we use the exponential decay model of the rabbit population, which is given by N(t) = a × e^(-kt). We know that N(0) = 232 rabbits five years ago and N(5) = 146 rabbits now. The first step is to solve for the constants a and k using the two given conditions.

Step 1: Find constant a using initial condition. We assume that t = 0 corresponds to five years ago, hence N(0) = 232 gives us a = 232.

Step 2: Find constant k using the current population. Plugging in N(5) = 146 and a = 232 into the model:
146 = 232 × e^(-5k)
Divide both sides by 232 to isolate the exponential term, and then take the natural logarithm to solve for k:

e^(-5k) = 146/232
-5k = ln(146/232)
Step 3: Determine the time t when the population will be two (N(t)=2). Using the model with solved constants, the equation becomes 2 = 232 × e^(-k * t). Solve for t and round to the nearest year.

Conclusion: The solution will give us the number of years from now when only two rabbits will be left on the island.