After years of overhunting, environmental scientists have reintroduced wolverines to Yellowstone National Park. The initial number of wolverines reintroduced to the park was 88 and, after 5 years, the population is estimated to be around 578. Assuming an exponential growth pattern, what is the annual growth rate (rounded to the nearest tenth of a percent) of the new wolverine population in Yellowstone National Park?

Hint: A(t) = A0(1 + r)t, where A(t) is the final amount, A0 is the initial amount, r is the growth rate expressed as a decimal amount, and t is time.

Question

After years of overhunting, environmental scientists have reintroduced wolverines to Yellowstone National Park. The initial number of wolverines reintroduced to the park was 88 and, after 5 years, the population is estimated to be around 578. Assuming an exponential growth pattern, what is the annual growth rate (rounded to the nearest tenth of a percent) of the new wolverine population in Yellowstone National Park?

Hint: A(t) = A0(1 + r)t, where A(t) is the final amount, A0 is the initial amount, r is the growth rate expressed as a decimal amount, and t is time.

justjm · Answer

Apply the general formula for exponential growth

$ A(t) = A_{0}(1+r)^{t} $
You are given that $ A_0 $ is 88 and after $ t=5$ years, $A(t)=578$
Plug that in

$ 578 = 88(1+r)^5$
Solve for r
$ \frac{578}{88} = (1+r)^5 $
r=0.4567
So the annual growth rate is 45.67%