Question

the count in a bacteria culture was 300 after 10 minutes and 1500 after 35 minutes. assuming the count grows exponentially,
what was the initial size of the culture?
find the doubling period.
find the population after 105 minutes.
when will the population reach 10000.
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 \(P(10) = 300\), so \(300=P_0e^{10k}\), and \(P(35)=1500\), so \(1500 = P_0e^{35k}\).
Divide the second equation by the first equation: \(\frac{1500}{300}=\frac{P_0e^{35k}}{P_0e^{10k}}\).
Simplify the left - hand side to get \(5\), and the right - hand side simplifies to \(e^{25k}\) (using the property \(\frac{e^{a}}{e^{b}}=e^{a - b}\)).
So, \(e^{25k}=5\). Take the natural logarithm of both sides: \(25k=\ln(5)\), then \(k=\frac{\ln(5)}{25}\approx\frac{1.6094}{25}=0.0644\).
Substitute \(k\) into the equation \(300 = P_0e^{10k}\). We have \(300=P_0e^{10\times\frac{\ln(5)}{25}}\).
Simplify \(10\times\frac{\ln(5)}{25}=\frac{2\ln(5)}{5}=\ln(5^{\frac{2}{5}})\) (using the property \(a\ln(b)=\ln(b^{a})\)).
So, \(300 = P_0\times5^{\frac{2}{5}}\). Then \(P_0=\frac{300}{5^{\frac{2}{5}}}\). Since \(5^{\frac{2}{5}}=\sqrt[5]{25}\approx1.9037\), \(P_0=\frac{300}{1.9037}\approx157.69\).

Step2: Find the doubling period

For the doubling period \(T\), we use the formula \(P(t + T)=2P(t)\). Substituting into \(P(t)=P_0e^{kt}\), we get \(P_0e^{k(t + T)}=2P_0e^{kt}\).
Cancel out \(P_0e^{kt}\) (since \(P_0
eq0\) and \(e^{kt}
eq0\) for \(t\geq0\)), and we have \(e^{kT}=2\).
Take the natural logarithm of both sides: \(kT=\ln(2)\). Since \(k = \frac{\ln(5)}{25}\), then \(T=\frac{25\ln(2)}{\ln(5)}\).
Using \(\ln(2)\approx0.6931\) and \(\ln(5)\approx1.6094\), \(T=\frac{25\times0.6931}{1.6094}=\frac{17.3275}{1.6094}\approx10.76\) minutes.

Step3: Find the population after \(t = 105\) minutes

We know \(P_0\approx157.69\) and \(k=\frac{\ln(5)}{25}\). Using \(P(t)=P_0e^{kt}\), substitute \(t = 105\).
\(P(105)=157.69e^{105\times\frac{\ln(5)}{25}}\).
Simplify \(105\times\frac{\ln(5)}{25}=\frac{21\ln(5)}{5}=\ln(5^{\frac{21}{5}})\) (using \(a\ln(b)=\ln(b^{a})\)).
\(5^{\frac{21}{5}}=(5^{4})\times5^{\frac{1}{5}} = 625\times\sqrt[5]{5}\approx625\times1.3797\approx862.31\).
\(P(105)=157.69\times862.31\approx136000\) (using \(P(t)=P_0e^{kt}\) with \(P_0 = 157.69\) and \(k=\frac{\ln(5)}{25}\), \(P(105)=157.69e^{\frac{105\ln(5)}{25}}=157.69\times5^{\frac{105}{25}}=157.69\times5^{4.2}\), \(5^{4.2}=5^{4}\times5^{0.2}=625\times\sqrt[5]{5}\approx625\times1.3797\), \(P(105)=157.69\times(625\times1.3797)\approx136000\)).

Step4: Find when the population reaches \(P(t)=10000\)

Using \(P(t)=P_0e^{kt}\), substitute \(P(t) = 10000\), \(P_0\approx157.69\), and \(k=\frac{\ln(5)}{25}\).
\(10000=157.69e^{\frac{\ln(5)}{25}t}\).
First, \(\frac{10000}{157.69}=e^{\frac{\ln(5)}{25}t}\).
\(\frac{10000}{157.69}\approx63.42\). So, \(e^{\frac{\ln(5)}{25}t}=63.42\).
Take the natural logarithm of both sides: \(\frac{\ln(5)}{25}t=\ln(63.42)\).
Since \(\ln(63.42)\approx4.15\) and \(\ln(5)\approx1.6094\), \(t=\frac{25\times4.15}{1.6094}=\frac{103.75}{1.6094}\approx64.46\) minutes.

Answer:

• Initial size: \(157.69\)
• Doubling period: \(10.76\) minutes
• Population after \(105\) minutes: \(136000\)
• Time to reach \(10000\): \(64.46\) minutes