Question

this exercise uses the exponential growth model.
the bat population in a certain midwestern county was 260,000 in 2012, and the observed doubling time for the population is 32 years.
(a) find an exponential model ( n(t)=n_{0} 2^{t / a} ) for the population ( t ) years after 2012.
( n(t)= )
(b) find an exponential model ( n(t)=n_{0} e^{r t} ) for the population ( t ) years after 2012. (round your ( r ) value to four decimal places.)
( n(t)= )
(c) sketch a graph of the population at time ( t ).
(d) estimate how long (in yr) it takes the population to reach 2 million. (round your answer to two decimal places.)

Explanation:

Step1: Determine \(n_0\) and \(a\) for \(n(t)=n_02^{t/a}\)

Given \(n_0 = 260000\) (initial population in 2012) and \(a = 32\) (doubling - time). So the model is \(n(t)=260000\times2^{t/32}\).

Step2: Relate \(n(t)=n_0e^{rt}\) and \(n(t)=n_02^{t/a}\)

We know that \(n_0e^{rt}=n_02^{t/a}\). Take the natural logarithm of both sides: \(\ln(e^{rt})=\ln(2^{t/a})\). Using the property \(\ln(e^{x}) = x\) and \(\ln(a^{b})=b\ln(a)\), we get \(rt=\frac{t}{a}\ln(2)\). Solving for \(r\), when \(t
eq0\), \(r = \frac{\ln(2)}{a}\). Substituting \(a = 32\), \(r=\frac{\ln(2)}{32}\approx\frac{0.6931}{32}\approx0.0217\). So \(n(t)=260000e^{0.0217t}\).

Step3: Solve for \(t\) when \(n(t)=2000000\) using \(n(t)=n_0e^{rt}\)

Substitute \(n(t) = 2000000\), \(n_0=260000\) and \(r = 0.0217\) into \(n(t)=n_0e^{rt}\). We have \(2000000=260000e^{0.0217t}\). First, divide both sides by \(260000\): \(\frac{2000000}{260000}=e^{0.0217t}\), so \(\frac{100}{13}=e^{0.0217t}\). Then take the natural logarithm of both sides: \(\ln(\frac{100}{13})=\ln(e^{0.0217t})\). Using \(\ln(\frac{a}{b})=\ln(a)-\ln(b)\) and \(\ln(e^{x}) = x\), we get \(\ln(100)-\ln(13)=0.0217t\). Since \(\ln(100)\approx4.6052\) and \(\ln(13)\approx2.5649\), then \(4.6052 - 2.5649=0.0217t\). So \(2.0403 = 0.0217t\). Solving for \(t\), \(t=\frac{2.0403}{0.0217}\approx93.93\).

Answer:

(a) \(n(t)=260000\times2^{t/32}\)
(b) \(n(t)=260000e^{0.0217t}\)
(d) \(t\approx93.93\) yr