Questions: The count in a bacteria culture was 300 after 15 minutes and 1000 after 35 minutes. Assuming the count grows exponentially, What was the initial size of the culture? Find the doubling period. Find the population after 75 minutes. When will the population reach 10000. You may enter the exact value or round to 2 decimal places.

Solution Steps

To solve exponential growth problems, first determine the growth rate using given data points and the exponential growth formula. Then, use this rate to find the initial population size, doubling period, future population at a specific time, and the time it will take to reach a certain population size.

Using the exponential growth formula $P(t) = P_0 e^{kt}$, we have two data points:

1. $P_1 = 300$ at $t_1 = 15$
2. $P_2 = 1000$ at $t_2 = 35$

By dividing the equations, we eliminate $P_0$: $$\frac{1000}{300} = e^{20k} \implies \frac{10}{3} = e^{20k}$$ Taking the natural logarithm: $$\ln\left(\frac{10}{3}\right) = 20k \implies k = \frac{\ln\left(\frac{10}{3}\right)}{20} \approx 0.060198640216296805$$

Now, substituting $k$ back into the equation for $P_1$: $$P_0 = \frac{P_1}{e^{kt_1}} = \frac{300}{e^{15k}} \approx 121.6080139326331$$ Thus, the initial size of the culture is: $$\boxed{P_0 \approx 121.61}$$

The doubling period $T$ can be calculated using the formula: $$T = \frac{\ln(2)}{k} \approx 11.514332849868898$$ Thus, the doubling period is: $$\boxed{T \approx 11.51 \text{ minutes}}$$

To find the population after $t_3 = 75$ minutes, we use: $$P(75) = P_0 e^{k \cdot 75} \approx 11111.111111111117$$ Thus, the population after 75 minutes is: $$\boxed{P(75) \approx 11111.11}$$

Final Answer