a 5% growth per year, starting at 18000 means n(t) = 18000 * 1.05^t
Since 1.05 = e^(ln 1.05) that makes
n = 18000 (e^(ln 1.05))^t
n = 18000 e^(ln 1.05 * t)
n = 18000 e^(.0488t)
Now you can answer the other questions.
The fox population in a certain region has a relative growth rate of 5% per year. It is estimated that the population in 2013 was 18,000.
(a) Find a function
n(t) = n0ert
that models the population t years after 2013.
n(t) =
Correct: Your answer is correct.
(b) Use the function from part (a) to estimate the fox population in the year 2021. (Round your answer to the nearest whole number.)
26853
Correct: Your answer is correct.
foxes
(c) After how many years will the fox population reach 27,000? (Round your answer to one decimal place.)
8
Incorrect: Your answer is incorrect.
yr
(d) Sketch a graph of the fox population function for the years 2013–2021.
2 answers
looks like you missed d)
18000 e^(0.05t) = 27000
e^(.05t) = 1.5
take ln of both sides and use log rules
.05t = ln1.5
t = ln1.5/.05 = appr 8.1 years past 2013
18000 e^(0.05t) = 27000
e^(.05t) = 1.5
take ln of both sides and use log rules
.05t = ln1.5
t = ln1.5/.05 = appr 8.1 years past 2013