Question

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 ).

Explanation:

Step1: Find the value of \(n_0\) and \(a\) for \(n(t)=n_02^{t/a}\)

Given that in 2012 (\(t = 0\)), the population \(n(0)=n_0 = 260000\). The doubling - time formula for \(n(t)=n_02^{t/a}\): when the population doubles, \(n(t)=2n_0\). So, \(2n_0=n_02^{t/a}\), which simplifies to \(2 = 2^{t/a}\). Since the doubling - time \(t = 32\) years, then \(a = 32\). So, \(n(t)=260000\times2^{t/32}\).

Step2: Find the value of \(r\) for \(n(t)=n_0e^{rt}\)

We know that when \(t = 32\), \(n(32)=2n_0\). Substitute into \(n(t)=n_0e^{rt}\): \(2n_0=n_0e^{32r}\). Divide both sides by \(n_0\) (since \(n_0
eq0\)), we get \(2 = e^{32r}\). Take the natural logarithm of both sides: \(\ln(2)=\ln(e^{32r})\). Using the property \(\ln(e^{x})=x\), we have \(\ln(2)=32r\). Solve for \(r\): \(r=\frac{\ln(2)}{32}\approx\frac{0.6931}{32}\approx0.0217\). So, \(n(t)=260000e^{0.0217t}\)

Answer:

(a) \(n(t)=260000\times2^{t/32}\)
(b) \(n(t)=260000e^{0.0217t}\)