Question

a population of e. coli bacteria grows exponentially with time. you believe that the mean time between divisions is ( t_{b}=32) min, and that cell death occurs on average after ( t_{m}=200) hr. the population starts with 7000 cells. complete parts (a) through (c).
(a) use the equation ( n(t)=n_{0}e^{(b - m)t}) to predict how many cells are present after 5 hours.
the number of cells present after 5 hours is (square).
(do not round until the final answer. then round to the nearest integer as needed.)

Explanation:

Step1: Find the value of \(b - m\)

The formula for exponential growth of bacteria is \(N(t)=N_{0}e^{(b - m)t}\). The mean - time between divisions \(t_{b}=32\) min. Since \(t\) is in hours, \(t_{b}=\frac{32}{60}\) h. For exponential growth, if \(N(t)=2N_{0}\) (when the population doubles), then \(2N_{0}=N_{0}e^{(b - m)t_{b}}\). Canceling \(N_{0}\), we get \(2 = e^{(b - m)t_{b}}\). Taking the natural logarithm of both sides: \(\ln(2)=(b - m)t_{b}\), so \(b - m=\frac{\ln(2)}{t_{b}}\). Substituting \(t_{b}=\frac{32}{60}\) h, we have \(b - m=\frac{\ln(2)}{\frac{32}{60}}=\frac{60\ln(2)}{32}=\frac{15\ln(2)}{8}\approx\frac{15\times0.6931}{8}\approx1.30\).

Step2: Substitute into the formula

We know \(N_{0} = 7000\) (initial population) and \(t = 5\) h. Using the formula \(N(t)=N_{0}e^{(b - m)t}\), substitute \(N_{0}=7000\), \(b - m=\frac{15\ln(2)}{8}\), and \(t = 5\). Then \(N(5)=7000e^{\frac{15\ln(2)}{8}\times5}\). Since \(a\ln(b)=\ln(b^{a})\), \(\frac{15\ln(2)}{8}\times5=\ln(2^{\frac{75}{8}})\). So \(N(5)=7000\times2^{\frac{75}{8}}\). \(2^{\frac{75}{8}}=2^{9+\frac{3}{8}}=2^{9}\times2^{\frac{3}{8}}\approx512\times1.296\). Then \(N(5)=7000\times512\times1.296\approx7000\times666.552\approx4665864\).

Answer:

\(4665864\)