Question

the count in a bacteria culture was 300 after 15 minutes and 1800 after 40 minutes. assuming the count grows exponentially,
what was the initial size of the culture?
find the doubling period.
find the population after 95 minutes.
when will the population reach 14000.
you may enter the exact value or round to 2 decimal places.

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 \(P(15) = 300\), so \(300=P_0e^{15k}\) (Equation 1), and \(P(40)=1800\), so \(1800 = P_0e^{40k}\) (Equation 2).
Divide Equation 2 by Equation 1: \(\frac{1800}{300}=\frac{P_0e^{40k}}{P_0e^{15k}}\).
Simplify to get \(6=e^{25k}\).
Take the natural logarithm of both sides: \(\ln(6)=25k\), so \(k=\frac{\ln(6)}{25}\approx0.0716\).
Substitute \(k\) into Equation 1: \(300 = P_0e^{15\times\frac{\ln(6)}{25}}\).
\(300=P_0e^{\frac{3\ln(6)}{5}}\), and since \(e^{\frac{3\ln(6)}{5}}=6^{\frac{3}{5}}\approx4.647\), then \(P_0=\frac{300}{6^{\frac{3}{5}}}\approx64.55\).

Step2: Find the doubling period

For doubling, \(P(t) = 2P_0\). Using \(P(t)=P_0e^{kt}\), we have \(2P_0=P_0e^{kt}\), so \(2 = e^{kt}\).
Take the natural - logarithm: \(\ln(2)=kt\). Since \(k = \frac{\ln(6)}{25}\), then \(t=\frac{25\ln(2)}{\ln(6)}\approx9.78\) minutes.

Step3: Find the population after 95 minutes

Using \(P(t)=P_0e^{kt}\), with \(P_0\approx64.55\), \(k=\frac{\ln(6)}{25}\), and \(t = 95\).
\(P(95)=64.55e^{\frac{\ln(6)}{25}\times95}\).
Since \(\frac{\ln(6)}{25}\times95=\frac{19\ln(6)}{5}=\ln(6^{\frac{19}{5}})\), then \(P(95)=64.55\times6^{\frac{19}{5}}\approx64.55\times46656^{\frac{1}{5}}\approx64.55\times6.866\approx443.54\).

Step4: Find when the population reaches 14000

Set \(P(t)=14000\), so \(14000 = 64.55e^{\frac{\ln(6)}{25}t}\).
\(\frac{14000}{64.55}=e^{\frac{\ln(6)}{25}t}\).
Take the natural - logarithm: \(\ln(\frac{14000}{64.55})=\frac{\ln(6)}{25}t\).
\(t=\frac{25\ln(\frac{14000}{64.55})}{\ln(6)}\approx\frac{25\times4.477}{1.792}\approx62.67\) minutes.

Answer:

Initial size: \(\approx64.55\)
Doubling period: \(\approx9.78\) minutes
Population after 95 minutes: \(\approx443.54\)
Time to reach 14000: \(\approx62.67\) minutes