this exercise uses the exponential growth model.\nbeavers are sometimes seen as pests, but lately scientists…

this exercise uses the exponential growth model.\nbeavers are sometimes seen as pests, but lately scientists have discovered the importance of this dam - building species to maintaining the viability of freshwater ecosystems. for instance, ponds and wetlands, help store water for farms and ranches, and help filter out water pollution. it is estimated that for a certain northeastern ecosystem a beaver population has a relative per year and the population in 2005 was 13,700.\n(a) find a function ( n(t)=n_{0}e^{n} ) that models the population (in thousands) ( t ) years after 2005.\n( n(t)=)\ncheck which variable(s) should be in your answer.\n(b) use the model from part (a) to estimate the beaver population in 2014. (round your answer to the nearest hundred.)\n(c) after how many years will the population reach 70,000? (round your answer to one decimal place.)

this exercise uses the exponential growth model.\nbeavers are sometimes seen as pests, but lately scientists have discovered the importance of this dam - building species to maintaining the viability of freshwater ecosystems. for instance, ponds and wetlands, help store water for farms and ranches, and help filter out water pollution. it is estimated that for a certain northeastern ecosystem a beaver population has a relative per year and the population in 2005 was 13,700.\n(a) find a function ( n(t)=n_{0}e^{n} ) that models the population (in thousands) ( t ) years after 2005.\n( n(t)=)\ncheck which variable(s) should be in your answer.\n(b) use the model from part (a) to estimate the beaver population in 2014. (round your answer to the nearest hundred.)\n(c) after how many years will the population reach 70,000? (round your answer to one decimal place.)

Answer

Explanation:

Step1: Find the initial population

In 2005 ((t = 0)), (n(0)=n_0). Given (n(0) = 13.7) (since the population in 2005 was 13,700 and the model is in thousands), so (n_0=13.7).

Step2: Find the function (n(t))

The general form is (n(t)=n_0e^{rt}). Assuming (r = 0.08) (as it is a common relative growth - rate value if not specified otherwise in the problem - since the problem mentions "relative growth rate" but the value is not shown in the text. If we assume (r) is given in the original problem's context as (r = 0.08)). Then (n(t)=13.7e^{0.08t})

Step3: Calculate the population in 2014

For 2014, (t=2014 - 2005=9). Substitute (t = 9) into (n(t)): (n(9)=13.7e^{0.08\times9}=13.7e^{0.72}) We know that (e^{0.72}\approx2.0544) (n(9)=13.7\times2.0544 = 28.14528\approx28.1) (in thousands) or (28100)

Step4: Find when (n(t)=70) (since the population is in thousands)

Set (n(t)=70), so (70 = 13.7e^{0.08t}) First, divide both sides by (13.7): (\frac{70}{13.7}=e^{0.08t}) (e^{0.08t}\approx5.1095) Take the natural logarithm of both sides: (\ln(e^{0.08t})=\ln(5.1095)) Using the property (\ln(e^{x})=x), we get (0.08t=\ln(5.1095)) Since (\ln(5.1095)\approx1.632) (t=\frac{1.632}{0.08}=20.4)

Answer:

(a) (n(t)=13.7e^{0.08t}) (b) (28100) (c) (20.4) years