Question

the count in a bacteria culture was 500 after 10 minutes and 1900 after 40 minutes. assuming the count grows exponentially,
what was the initial size of the culture?
find the doubling period.
find the population after 100 minutes.
when will the population reach 12000.
you may enter the exact value or round to 2 decimal places.
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Explanation:

Step1: Define exponential growth model

The exponential growth formula is $N(t) = N_0 e^{kt}$, where $N(t)$ is the population at time $t$, $N_0$ is the initial population, and $k$ is the growth rate.

Step2: Set up equations from given data

For $t=10$, $N(10)=500$: $500 = N_0 e^{10k}$
For $t=40$, $N(40)=1900$: $1900 = N_0 e^{40k}$

Step3: Solve for $k$

Divide the second equation by the first:
$\frac{1900}{500} = \frac{N_0 e^{40k}}{N_0 e^{10k}}$
$3.8 = e^{30k}$
Take natural log of both sides:
$\ln(3.8) = 30k$
$k = \frac{\ln(3.8)}{30} \approx \frac{1.3350}{30} \approx 0.0445$

Step4: Solve for initial population $N_0$

Substitute $k$ into $500 = N_0 e^{10k}$:
$N_0 = \frac{500}{e^{10 \times 0.0445}} = \frac{500}{e^{0.445}} \approx \frac{500}{1.560} \approx 320.51$

Step5: Find population at 100 minutes

Use $N(100) = N_0 e^{100k}$:
$N(100) \approx 320.51 e^{100 \times 0.0445} = 320.51 e^{4.45} \approx 320.51 \times 85.74 \approx 27480.53$

Step6: Find time to reach 12000

Set $N(t)=12000$ and solve for $t$:
$12000 = 320.51 e^{0.0445t}$
$\frac{12000}{320.51} = e^{0.0445t}$
$37.44 = e^{0.0445t}$
Take natural log:
$\ln(37.44) = 0.0445t$
$t = \frac{\ln(37.44)}{0.0445} \approx \frac{3.623}{0.0445} \approx 81.42$

Answer:

Initial size of the culture: $\approx 320.51$
Population after 100 minutes: $\approx 27480.53$
Time to reach 12000: $\approx 81.42$ minutes