Question

the count in a bacteria culture was 800 after 20 minutes and 1500 after 40 minutes. assuming the count grows exponentially,
what was the initial size of the culture?
find the doubling period.
find the population after 90 minutes.
when will the population reach 10000.
you may enter the exact value or round to 2 decimal places.
question help: video read written example

Explanation:

Step1: Set up the exponential growth formula

The exponential growth formula is \(P(t)=P_0e^{kt}\), where \(P(t)\) is the population at time \(t\), \(P_0\) is the initial population, and \(k\) is the growth constant.
We know that \(P(20) = 800\) and \(P(40)=1500\).
Substituting \(t = 20\) into the formula: \(800=P_0e^{20k}\) (Equation 1)
Substituting \(t = 40\) into the formula: \(1500=P_0e^{40k}\) (Equation 2)

Step2: Solve for \(k\)

Divide Equation 2 by Equation 1:
\(\frac{1500}{800}=\frac{P_0e^{40k}}{P_0e^{20k}}\)
\(\frac{15}{8}=e^{20k}\)
Take the natural logarithm of both sides:
\(\ln(\frac{15}{8}) = 20k\)
\(k=\frac{\ln(\frac{15}{8})}{20}\approx\frac{0.6286}{20}=0.03143\)

Step3: Solve for \(P_0\)

Substitute \(k\) into Equation 1:
\(800 = P_0e^{20\times0.03143}\)
\(800 = P_0e^{0.6286}\)
\(P_0=\frac{800}{e^{0.6286}}\approx\frac{800}{1.872}\approx427.35\)

Step4: Find the doubling period

The doubling period \(T\) satisfies \(2P_0=P_0e^{kT}\)
\(2 = e^{kT}\)
Take the natural logarithm of both sides: \(\ln(2)=kT\)
\(T=\frac{\ln(2)}{k}=\frac{\ln(2)}{\frac{\ln(\frac{15}{8})}{20}}=\frac{20\ln(2)}{\ln(\frac{15}{8})}\approx\frac{20\times0.6931}{0.6286}\approx22.06\) minutes

Step5: Find the population after \(t = 90\) minutes

\(P(90)=P_0e^{90k}\)
Substitute \(P_0\approx427.35\) and \(k\approx0.03143\)
\(P(90)=427.35e^{90\times0.03143}\)
\(P(90)=427.35e^{2.8287}\)
\(P(90)=427.35\times16.91\approx7230.00\)

Step6: Find when the population reaches \(P(t) = 10000\)

\(10000=427.35e^{0.03143t}\)
\(\frac{10000}{427.35}=e^{0.03143t}\)
\(\ln(\frac{10000}{427.35})=0.03143t\)
\(t=\frac{\ln(\frac{10000}{427.35})}{0.03143}\approx\frac{3.167}{0.03143}\approx100.76\) minutes

Answer:

Initial size: \(427.35\)
Doubling period: \(22.06\) minutes
Population after \(90\) minutes: \(7230.00\)
Time to reach \(10000\): \(100.76\) minutes