Question

After a shipwreck, 120 rats manage to swim from the wreckage to a deserted island. The rat population on the island grows exponentially, and after 15 months, there are 280 rats on the island.
A. Find a function that models the population t months after the arrival of the rats.
B. What will the population be 3 years after the shipwreck?
C. When will the population reach 2000 rats?

Answer 1

Answer:

a.   $X(T) = 120 (1.058)^T$

b.   Population after 3 years is 142

c.   50 years

Step-by-step explanation:

Given

Type of growth: Exponential

Initial number of rats = 120

Number of rats (15months) = 280

Solving (a)

Since the growth type is exponential, we make use of the following exponential progression

$X_T = X_0 (1 + R)^T$

Where Xo is the initial population;

Xo = 125

$X_T$ is the current population at T month

So;

$X_T = 280$; when $T = 15$

Substitute these values in the above formula

$280 =120 * (1 + R)^{15}$

Divide both sides by 120

$\frac{280}{120} =(1 + R)^{15}$

$2.3333 =(1 + R)^{15}$

Take 15th root of both sides

$\sqrt[15]{2.3333} =1 + R$

$1.05811235561 = 1 + R$

Subtract 1 from both sides

$R = 1.05811235561 - 1$

$R = 0.05811235561$

$R = 0.058$ (Approximated)

Plug in values of R and Xo in $X_T = X_0 (1 + R)^T$

$X_T = 120 (1 + 0.058)^T$

$X_T = 120 (1.058)^T$

Write as a function

$X(T) = 120 (1.058)^T$

Hence, the function is $X(T) = 120 (1.058)^T$

Solving (b):

Population after 3 years

In this case, T = 3

So:

$X(T) = 120 (1.058)^T$

$X(3) = 120 (1.058)^3$

$X(3) = 120 * 1.18466445254$

$X(3) = 142.159734305$

$X(3) = 142$ (Approximated)

Solving (c): When population will reach 2000

Here: X(T) = 2000

So:

So:

$2000 = 120 (1.058)^T$

Divide both sides by 120

$\frac{2000}{120} = 1.058^T$

$16.667 = 1.058^T$

Take Log of both sides

$Log(16.667) = Log(1.058^T)$

Apply law of logarithm

$Log(16.667) = TLog(1.058)$

Divide both sides by Log(1.058)

$T = \frac{Log(16.667)}{Log(1.058)}$

$T = 49.9009236926$

Approximate

$T = 50\ years$