The population of fish in a lake is reduced by 10% each week because of many people fishing there. Find the half-life of this fish population exactly in weeks

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? 

---

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.

The population of fish in a lake is reduced by 10% each week because of many people fishing there. Find the half-life of this fish population exactly in weeks.

Discover how to calculate the half-life of a fish population that decreases by 10% each week. This detailed solution covers exponential decay, natural logarithms, and provides step-by-step instructions for finding the half-life of approximately 6.58 weeks.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation