Question

the count in a bacteria culture was 500 after 20 minutes and 1700 after 35 minutes. assuming the count grows exponentially,
what was the initial size of the culture?
find the doubling period.
find the population after 60 minutes.
when will the population reach 10000.
you may enter the exact value or round to 2 decimal

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 \(P(20) = 500\), so \(500=P_0e^{20k}\), and \(P(35)=1700\), so \(1700 = P_0e^{35k}\).
Divide the second equation by the first equation: \(\frac{1700}{500}=\frac{P_0e^{35k}}{P_0e^{20k}}\).
Simplify the left - hand side to get \(3.4\), and the right - hand side simplifies to \(e^{15k}\) (using the property \(\frac{e^{a}}{e^{b}}=e^{a - b}\)).
Take the natural logarithm of both sides: \(\ln(3.4)=\ln(e^{15k})\).
Since \(\ln(e^{x})=x\), we have \(15k=\ln(3.4)\), so \(k=\frac{\ln(3.4)}{15}\approx\frac{1.2238}{15}\approx0.0816\).

Step2: Find the initial population \(P_0\)

Substitute \(k\) into the equation \(500 = P_0e^{20k}\).
\(P_0=\frac{500}{e^{20k}}\), substituting \(k = \frac{\ln(3.4)}{15}\), we get \(P_0=\frac{500}{e^{20\times\frac{\ln(3.4)}{15}}}\).
Using the property \(a\ln(b)=\ln(b^{a})\) and \(e^{\ln(x)}=x\), \(e^{20\times\frac{\ln(3.4)}{15}}=(3.4)^{\frac{20}{15}}=(3.4)^{\frac{4}{3}}\approx3.4^{1.333}\approx6.08\).
So \(P_0=\frac{500}{(3.4)^{\frac{4}{3}}}\approx\frac{500}{6.08}\approx82.24\).

Step3: Find the doubling period \(T\)

For doubling, \(P(T) = 2P_0\). Using the formula \(P(T)=P_0e^{kT}\), we have \(2P_0=P_0e^{kT}\).
Cancel out \(P_0\) (since \(P_0
eq0\)), and we get \(2 = e^{kT}\).
Take the natural logarithm of both sides: \(\ln(2)=\ln(e^{kT})\), so \(T=\frac{\ln(2)}{k}\).
Substituting \(k=\frac{\ln(3.4)}{15}\), we have \(T=\frac{15\ln(2)}{\ln(3.4)}\approx\frac{15\times0.6931}{1.2238}\approx8.5\) minutes.

Step4: Find the population after \(t = 60\) minutes

Using the formula \(P(t)=P_0e^{kt}\), with \(P_0\approx82.24\), \(k\approx0.0816\), and \(t = 60\).
\(P(60)=82.24e^{0.0816\times60}\).
\(0.0816\times60 = 4.896\), \(e^{4.896}\approx133.7\).
So \(P(60)\approx82.24\times133.7\approx10996.5\).

Step5: Find when the population reaches \(P(t)=10000\)

Set \(10000 = 82.24e^{0.0816t}\).
First, divide both sides by \(82.24\): \(\frac{10000}{82.24}=e^{0.0816t}\).
\(\frac{10000}{82.24}\approx121.6\).
Take the natural logarithm of both sides: \(\ln(121.6)=\ln(e^{0.0816t})\).
Since \(\ln(e^{x})=x\), \(t=\frac{\ln(121.6)}{0.0816}\).
\(\ln(121.6)\approx4.8\), so \(t=\frac{4.8}{0.0816}\approx58.8\) minutes.

Answer:

• Initial size: \(82.24\)
• Doubling period: \(8.5\) minutes
• Population after 60 minutes: \(10996.5\)
• Time to reach 10000: \(58.8\) minutes