Question

score on last try: 0.75 of 1 pts. see details for more.
at least one scored part is incorrect. jump to first changable incorrect part.
you can retry this question below
the count in a bacteria culture was 100 after 10 minutes and 1200 after 30 minutes. assuming the count grows exponentially,
what was the initial size of the culture? bacteria
find the doubling period. 5.58 minutes
find the population after 75 minutes. 321000 bacteria
when will the population reach 15000. 50.28 minutes

Explanation:

Step1: Define exponential growth model

The general form of exponential growth is \( P(t) = P_0 e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, and \( k \) is the growth rate.
We know that at \( t = 10 \), \( P(10)=100 \) and at \( t = 30 \), \( P(30)=1200 \).
So we have two equations:
\( 100 = P_0 e^{10k} \) (Equation 1)
\( 1200 = P_0 e^{30k} \) (Equation 2)

Step2: Divide Equation 2 by Equation 1

Divide Equation 2 by Equation 1: \( \frac{1200}{100}=\frac{P_0 e^{30k}}{P_0 e^{10k}} \)
Simplify: \( 12 = e^{20k} \)

Step3: Solve for \( k \)

Take the natural logarithm of both sides: \( \ln(12)=20k \)
So \( k=\frac{\ln(12)}{20}\approx\frac{2.4849}{20} = 0.124245 \)

Step4: Substitute \( k \) into Equation 1 to find \( P_0 \)

From Equation 1: \( 100 = P_0 e^{10\times0.124245} \)
Calculate \( e^{10\times0.124245}=e^{1.24245}\approx3.4641 \)
Then \( P_0=\frac{100}{3.4641}\approx28.87 \approx 29 \) (or more accurately, let's do it symbolically first. From \( 100 = P_0 e^{10k} \) and \( 12 = e^{20k}=(e^{10k})^2 \), so \( e^{10k}=\sqrt{12} = 2\sqrt{3}\approx3.4641 \). Then \( P_0=\frac{100}{2\sqrt{3}}=\frac{50}{\sqrt{3}}\approx28.87 \))

Answer:

The initial size of the culture is approximately \(\boldsymbol{29}\) (or more precisely \(\frac{50}{\sqrt{3}}\approx28.87\)) bacteria.