(5 pts) Assume that the number of bacteria follows an exponential growth model: $P(t)=P_{0} e^{k t}$. The count in the bacteria culture was 900 after 10 minutes and 1000 after 40 minutes.
(a) What was the initial size of the culture?
(b) Find the population after 85 minutes.
(c) How many minutes after the start of the experiment will the population reach 13000 ?

Step 1 ：Given two points on the exponential growth curve: (10, 900) and (40, 1000). We can use these to solve for the constants P0 and k in the exponential growth model \(P(t) = P0 * e^{kt}\).

Step 2 ：First, we'll solve for k using the two given points. We can create two equations from the points: \(900 = P0 * e^{10k}\) and \(1000 = P0 * e^{40k}\).

Step 3 ：Divide equation (2) by equation (1) to eliminate P0: \(\frac{1000}{900} = \frac{e^{40k}}{e^{10k}}\) which simplifies to \(1.111... = e^{30k}\).

Step 4 ：Take the natural logarithm of both sides to solve for k: \(\ln(1.111...) = 30k\) which gives \(k = \frac{\ln(1.111...)}{30} \approx 0.0036\).

Step 5 ：Now, substitute k into equation (1) to solve for P0: \(900 = P0 * e^{10 * 0.0036}\) which gives \(P0 = \frac{900}{e^{0.036}} \approx 868.6\).

Step 6 ：\(\boxed{P0 \approx 869}\) is the initial size of the culture.

Step 7 ：To find the population after 85 minutes, substitute t = 85, P0 = 869, and k = 0.0036 into the exponential growth model: \(P(85) = 869 * e^{0.0036 * 85} \approx 1200\).

Step 8 ：\(\boxed{P(85) \approx 1200}\) is the population after 85 minutes.

Step 9 ：To find out how many minutes after the start of the experiment will the population reach 13000, set \(P(t) = 13000\) and solve for t: \(13000 = 869 * e^{0.0036t}\) which gives \(t = \frac{\ln(13000 / 869)}{0.0036} \approx 460\).

Step 10 ：\(\boxed{t \approx 460}\) minutes after the start of the experiment, the population will reach 13000.