Question

under optimal conditions, the growth of a certain strain of e. coli is modeled by the law of uninhibited growth ( a(t)=a_{0} e^{k t} ) where ( a_{0} ) is the initial number of bacteria and ( t ) is the elapsed time, measured in minutes. from numerous experiments, it has been determined that the doubling time of this organism is 20 minutes. suppose 1900 bacteria are present initially.

a) find the exact value of ( k ), the growth constant. round to 6 decimal places or enter an exact answer.
( k = 0.034657 )

b) using the ( k ) value from part (a), find a function that gives the number of bacteria ( a(t) ) after ( t ) minutes.
( a(t)=1900 e^{0.034657 t} )

c) how long until there are 7200 bacteria? round to the nearest tenth of a minute.
time: 38.5 minutes

Explanation:

Step1: Substitute values into formula

Given \(A(t)=1900e^{0.034657t}\), and \(A(t) = 7200\). So, \(7200=1900e^{0.034657t}\).

Step2: Simplify the equation

Divide both sides by \(1900\): \(\frac{7200}{1900}=e^{0.034657t}\), which simplifies to \(\frac{72}{19}=e^{0.034657t}\).

Step3: Take natural logarithm

Take the natural logarithm of both sides: \(\ln(\frac{72}{19})=\ln(e^{0.034657t})\). Using the property \(\ln(e^{x}) = x\), we get \(\ln(\frac{72}{19})=0.034657t\).

Step4: Solve for \(t\)

Calculate \(\ln(\frac{72}{19})\approx1.3699\). Then \(t=\frac{\ln(\frac{72}{19})}{0.034657}\approx\frac{1.3699}{0.034657}\approx39.5\).

Answer:

\(39.5\) minutes