A population of plants grows according to the model P = A * (2^kt). Initially, there are 400 plants. After 2 years, there are 1600 plants. Calculate the growth rate k and the time t when the populatio

The population model is given by $P = A \cdot (2^{kt})$.
At the start of the project ($t=0$), there were 400 plants. Substituting this into the model:

$$400 = A \cdot (2^{k \cdot 0})$$ $$400 = A \cdot (2^0)$$ $$400 = A \cdot 1$$ $$A = 400$$

By the end of the second year ($t=2$), the population had increased to 1,600 plants. Using $A=400$:

$$1600 = 400 \cdot (2^{k \cdot 2})$$ $$1600 = 400 \cdot (2^{2k})$$

Divide both sides by 400:

$$\frac{1600}{400} = 2^{2k}$$ $$4 = 2^{2k}$$

Since $4 = 2^2$, we can write:

$$2^2 = 2^{2k}$$

Equating the exponents:

$$2 = 2k$$ $$k = \frac{2}{2}$$ $$k = 1$$

The growth rate $k$ is 1. This confirms the growth is consistent with the model.

We use the population model $P = A \cdot (2^{kt})$ with $A=400$ and $k=1$. We want to find $t$ when $P=25,600$.

$$25600 = 400 \cdot (2^{1 \cdot t})$$ $$25600 = 400 \cdot (2^t)$$

Divide both sides by 400:

$$\frac{25600}{400} = 2^t$$ $$64 = 2^t$$

Since $2^6 = 64$, we have:

$$2^6 = 2^t$$

Equating the exponents:

$$t = 6$$

The population will reach a self-sustaining level in 6 years.

The Stability Index formula is $S = \log_2 P - \log_2 A$.
For the self-sustaining level, $P = 25,600$ and $A = 400$.

$$S = \log_2 (25600) - \log_2 (400)$$

Using the logarithm property $\log_b x - \log_b y = \log_b \left(\frac{x}{y}\right)$:

$$S = \log_2 \left(\frac{25600}{400}\right)$$ $$S = \log_2 (64)$$

Since $2^6 = 64$, then $\log_2 (64) = 6$.

$$S = 6$$

The Stability Index at the self-sustaining level is 6.

The population will reach a self-sustaining level of 25,600 plants in $\boxed{\text{6 years}}$.
The Stability Index ($S$) at this level is $\boxed{6}$.