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 1200 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 = )
b) using the ( k ) value from part (a), find a function that gives the number of bacteria ( a(t) ) after ( t ) minutes.
( a(t) = )
c) how long until there are 9300 bacteria? round to the nearest tenth of a minute.
time: minutes
check answer

Explanation:

Step1: Find the value of \( k \)

Given \( A(t)=A_0e^{kt} \), when \( t = 20 \), \( A(20)=2A_0 \). Substitute into the formula:
\( 2A_0=A_0e^{20k} \). Divide both sides by \( A_0 \) (since \( A_0
eq0 \)): \( 2 = e^{20k} \).
Take the natural logarithm of both sides: \( \ln(2)=\ln(e^{20k}) \).
Using the property \( \ln(e^{x})=x \), we get \( \ln(2) = 20k \).
Solve for \( k \): \( k=\frac{\ln(2)}{20}\approx\frac{0.693147}{20}=0.034657 \).

Step2: Write the function \( A(t) \)

Since \( A_0 = 1200 \) and \( k=\frac{\ln(2)}{20} \), substitute into \( A(t)=A_0e^{kt} \).
\( A(t)=1200e^{\frac{\ln(2)}{20}t}=1200\times2^{\frac{t}{20}} \) (using \( a^{bc}=(a^{b})^{c} \) and \( e^{\ln(2)x}=2^{x} \)).

Step3: Find the time \( t \) when \( A(t) = 9300 \)

Set \( A(t)=9300 \), so \( 9300=1200e^{\frac{\ln(2)}{20}t} \).
Divide both sides by \( 1200 \): \( \frac{9300}{1200}=e^{\frac{\ln(2)}{20}t} \), which simplifies to \( \frac{31}{4}=e^{\frac{\ln(2)}{20}t} \).
Take the natural logarithm of both sides: \( \ln(\frac{31}{4})=\frac{\ln(2)}{20}t \).
Solve for \( t \): \( t=\frac{20\ln(\frac{31}{4})}{\ln(2)} \).
Calculate \( \ln(\frac{31}{4})\approx\ln(7.75)\approx2.047 \), \( \ln(2)\approx0.693 \).
\( t=\frac{20\times2.047}{0.693}\approx\frac{40.94}{0.693}\approx59.1 \).

Answer:

a) \( k = 0.034657 \)
b) \( A(t)=1200e^{0.034657t} \)
c) \( 59.1 \) minutes