To find the half-life of the fish population, we can start by using the exponential decay formula, which is:

$P(t) = P_0 \cdot e^{kt}$

where:

• $P_0$ is the initial population,
• $P(t)$ is the population at time $t$,
• $k$ is the decay constant (which we need to find),
• $t$ is time in weeks.

Step 1: Determine the decay constant $k$

Since the population reduces by 10% each week, after one week the population will be 90% of the previous week's population. Therefore,

$P(1) = 0.9 \cdot P_0$

Using the decay formula, we substitute $t = 1$ and $P(1) = 0.9 \cdot P_0$:

$0.9 \cdot P_0 = P_0 \cdot e^{k \cdot 1}$

Dividing both sides by $P_0$:

$0.9 = e^k$

Taking the natural logarithm of both sides to solve for $k$:

$k = \ln(0.9)$

Calculating this gives:

$k \approx -0.10536$

Step 2: Determine the half-life

The half-life is the time it takes for the population to reduce to half of its initial size. We set $P(t) = \frac{1}{2} P_0$ and solve for $t$:

$\frac{1}{2} P_0 = P_0 \cdot e^{kt}$

Dividing both sides by $P_0$:

$\frac{1}{2} = e^{kt}$

Taking the natural logarithm of both sides:

$\ln\left(\frac{1}{2}\right) = kt$

We know $\ln\left(\frac{1}{2}\right) = -\ln(2)$, so:

$-\ln(2) = -0.10536 \cdot t$

Solving for $t$:

$t = \frac{\ln(2)}{0.10536}$

Calculating this gives:

$t \approx 6.58 \text{ weeks}$

Answer:

The half-life of the fish population is approximately $6.58$ weeks.

Would you like more details or have any questions?

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1. How is exponential decay used in real-life population studies?
2. What factors influence the decay constant in such problems?
3. How does the half-life change with different decay rates?
4. What would happen if the decay rate were reduced from 10% to 5%?
5. How does the formula for half-life change in non-exponential decay situations?

Tip: The half-life formula for exponential decay is always $t = \frac{\ln(2)}{k}$, where $k$ is the decay constant.