Hello there. To solve this question, we need to pay attention to the data given by the question and set up an equation for the amount of rabbits in that colony.
In the year 2000 there were 1200 rabits in a colony, and it was observed that their population was increasing at a rate of 21% each year
For this type of question, we call f(x) the function that shows the population in a certain time x.
This function varies directly to the initial population and grows exponentially according to the rate .
Thus, we have that f(x) = P0 * (1+r)^x
In this case, P0 = 1200 and r = 0.21 (21% converted into decimals)
f(x) = 1200 * (1 + 0.21)^x
f(x) = 1200 * 1.21^x
This is the function that models the population of this colony for a year x.
To find how many rabbits you'll have in the year 2007, plug in x = 7
f(7) = 1200 * 1.21^7
Calculate the value
f(7) = 1200 * 3.797 = 4556
In which year will the population be equal to 5000
Making f(x) = 5000, we solve for x
5000 = 1200 * 1.21^x
Divide both sides of the equation by 1200
25/6 = 1.21^x
Take the natural log on both sides of the equation
ln(25/6) = ln(1.21^x)
Apply the logarithm power rule: log(a^b) = b * log(a)
ln(25/6) = x * ln(1.21)
Rewriting 1.21 as 121/100 = (11/10)², we get:
ln(25/6) = 2x * ln(11/10)
Divide both sides of the equation by a factor of 2ln(11/10)
x = ln(25/6)/(2ln(11/10)) approx. 7.4867 years
Rounding up to the next year, x = 8 years.