SOLUTION: In 2004, there were 3000 iguanas on Galapagos island. Since then, the population of Iguana on the island has increased by 5% each year. In what year will the population first excee

Algebra ->  Sequences-and-series -> SOLUTION: In 2004, there were 3000 iguanas on Galapagos island. Since then, the population of Iguana on the island has increased by 5% each year. In what year will the population first excee      Log On


   



Question 1165703: In 2004, there were 3000 iguanas on Galapagos island. Since then, the population of Iguana on the island has increased by 5% each year. In what year will the population first exceed 10,000?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
in 2004 there are 3000 iguanas.
the population grows by 5% per year.
when will the population exceed 10,000?

formula to use if f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the rate of increase each year.
n is the number of years.

formula becomes 10,000 = 3,000 * 1.05 ^ n

that's because r is equal to 5% per year / 100 = .05 per year.
1 + r is equal to 1 + .05 = 1.05

divide both sides of 10,000 = 3,000 * 1.05 ^ n by 3000 and simplify to get:
(10/3) = 1.05 ^ n
take the log of both sides of the equation to get:
log(10/3) = log(1.05 ^ n)
by log rules, this becomes:
log(10/3) = n * log(1.05)
divide both sides of the equation by log(1.05) to get:
log(10/3) / log(1.05) = n
solve for n to get:
n = log(10/3) / log(1.05) = 24.67654751.
that's the number of years when the population will be equal to 10,000.
since:
x = 0 is the beginning of the first year.
x = 1 is the end of the first year.
x = 24 is the end of the twenty-fourth year.
x = 25 is the end of the twenty-fifth year.
then:
the population will exceed 10,000 sometime between the end of the twenty-fourth year and the end of the twenty-fifth year.
that puts is in the 25th year.

here's a graph.



the points are in (x,y) format.
x is the end of the year indicated.
y is the population.