Question

the population of rabbits on an island is growing exponentially. in the year 2000, the population of rabbits was 100, and by 2007 the population had grown to 190. predict the population of rabbits in the year 2013, to the nearest whole number.

Explanation:

Step1: Recall the exponential growth formula

The general form of an exponential growth model is \( P(t) = P_0 e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( k \) is the growth rate constant, and \( t \) is the time elapsed.

Step2: Determine the time values

Let \( t = 0 \) correspond to the year 2000. Then, in 2007, \( t = 2007 - 2000 = 7 \) years, and in 2013, \( t = 2013 - 2000 = 13 \) years. We know that \( P_0 = 100 \) (population in 2000) and \( P(7) = 190 \) (population in 2007).

Step3: Solve for the growth rate \( k \)

Substitute \( P_0 = 100 \), \( t = 7 \), and \( P(7) = 190 \) into the exponential growth formula:

$$ 190 = 100 e^{k \cdot 7} $$

First, divide both sides by 100:

$$ \frac{190}{100} = e^{7k} $$

Simplify \( \frac{190}{100} = 1.9 \), so:

$$ 1.9 = e^{7k} $$

Take the natural logarithm of both sides:

$$ \ln(1.9) = \ln(e^{7k}) $$

Using the property \( \ln(e^x) = x \), we get:

$$ \ln(1.9) = 7k $$

Now, solve for \( k \) by dividing both sides by 7:

$$ k = \frac{\ln(1.9)}{7} $$

Calculate \( \ln(1.9) \approx 0.6419 \), so:

$$ k \approx \frac{0.6419}{7} \approx 0.0917 $$

Step4: Predict the population in 2013

Now that we have \( k \approx 0.0917 \), we can find \( P(13) \) (population in 2013) using the exponential growth formula with \( P_0 = 100 \), \( t = 13 \), and \( k \approx 0.0917 \):

$$ P(13) = 100 e^{0.0917 \cdot 13} $$

First, calculate the exponent: \( 0.0917 \cdot 13 \approx 1.1921 \)
Then, calculate \( e^{1.1921} \approx 3.295 \)
Multiply by 100:

$$ P(13) \approx 100 \cdot 3.295 = 329.5 $$

Rounding to the nearest whole number, we get \( P(13) \approx 330 \).

Answer:

330