(1) Bacterial Culture, growing exponentially, increase from ten hours - How much was present after three hours. (2) A bacterial Culture, growing exponentially, increase from 200 to 500 grams in the period from 6 am to 9 am. How many grams will be present at noon.

The exponential growth formula is $P(t) = P_0 e^{kt}$, where $P(t)$ is the amount at time $t$, $P_0$ is the initial amount, $k$ is the growth constant, and $t$ is the time.
Given:

• Initial amount $P_0 = 200$ grams at $t=0$ (6 am).
• Amount at 9 am is $P(3) = 500$ grams. The time elapsed from 6 am to 9 am is 3 hours.
• We need to find the amount at noon, which is $t=6$ hours from 6 am.

Substitute the known values into the formula:

$$500 = 200 e^{k \cdot 3}$$

Divide both sides by 200:

$$\frac{500}{200} = e^{3k}$$ $$2.5 = e^{3k}$$

Take the natural logarithm of both sides:

$$\ln(2.5) = \ln(e^{3k})$$ $$\ln(2.5) = 3k$$

Solve for $k$:

$$k = \frac{\ln(2.5)}{3}$$

Noon is 6 hours after 6 am ($t=6$). Use the calculated $k$ and the initial amount $P_0 = 200$:

$$P(6) = 200 e^{\left(\frac{\ln(2.5)}{3}\right) \cdot 6}$$ $$P(6) = 200 e^{2 \ln(2.5)}$$

Using the logarithm property $a \ln(b) = \ln(b^a)$:

$$P(6) = 200 e^{\ln((2.5)^2)}$$

Using the property $e^{\ln(x)} = x$:

$$P(6) = 200 \cdot (2.5)^2$$ $$P(6) = 200 \cdot 6.25$$ $$P(6) = 1250$$

The amount of bacterial culture present at noon will be $\boxed{\text{1250 grams}}$.