Question

A population of kangaroos is growing at a rate of 2% per year, compounded continuously. If the growth rate continues, how many years will it take for the size of the population to reach 150% of its current size according to the exponential growth function? Round your answer up to the nearest whole number, and do not include units.

Answer 1

Answer:

The correct answer is 20 years.

Explanation:

It is given that the growing rate is 2% or 2/100 = 0.02

This rate got increased to 150%

The formula for exponential growth is:  

A = Pe^rt

Let the initial population be 100%

150 = 100.e^0.02*t

3/2 = e^0.02t

1.5 = e^0.02t

After taking log from both the sides:  

ln(1.5) = 0.02t * ln(e) [ln(e) = 1]

ln(1.5) = 0.02t

t = ln(1.5)/0.02

t = 20.27

Thus, it will take around 20 years for the size of the population to reach 150 percent of its present size on the basis of the exponential growth function.  

Answer 2

Answer:

It will take 20 years for the size of the population to reach 150% of its current size according to the exponential growth function.

Explanation:

The exponential growth function is given by:

$P(t) = P(0)e^{rt}$

In which $P(t)$ is the population after t years, $P(0)$ is the initial population, e is the Euler number and r is the growth rate(decimal).

In this problem, we have that:

A population of kangaroos is growing at a rate of 2% per year, compounded continuously. This means that $r = 0.02$.

If the growth rate continues, how many years will it take for the size of the population to reach 150% of its current size according to the exponential growth function?

This is the value of t when $P(t) = 1.5P(0)$. So

$P(t) = P(0)e^{rt}$

$1.5P(0) = P(0)e^{0.02t}$

$e^{0.02t} = 1.5$

Since ln and e are inverse operation, we apply e to both sides of the equation to find t.

$\ln{e^{0.02t} }= \ln{1.5}$

$0.02t = 0.4055$

$t = \frac{0.4055}{0.02}$

$t = 20.27$

Rouding to the nearest whole number, it is 20 years.

It will take 20 years for the size of the population to reach 150% of its current size according to the exponential growth function.