Answer :
To find how many years it takes for the population of insects to double when growing at a constant rate, we need to iteratively apply the given growth factor until the population reaches twice its initial size.
Let's break this down step-by-step:
1. Given Values:
- The growth factor [tex]\( k \)[/tex] is 1.13.
- The initial population [tex]\( P_n \)[/tex] is 1 (for simplicity, we'll start with 1 unit).
- The target population is 2, which is double the initial population.
2. Initialize Variables:
- Start with the initial population, [tex]\( P_n = 1 \)[/tex].
- We need a counter for the number of years, starting at 0.
3. Growth Process:
- Each year, the population grows by multiplying the current population by the growth factor [tex]\( k = 1.13 \)[/tex].
- We keep track of how many years this process takes until the population reaches or exceeds the target.
Let's describe this process step-by-step:
1. Year 0:
- Population: [tex]\( P_0 = 1 \)[/tex]
- Target: 2
2. Year 1:
- New Population: [tex]\( P_1 = P_0 \times 1.13 = 1 \times 1.13 = 1.13 \)[/tex]
- Continue to next year as [tex]\( 1.13 < 2 \)[/tex]
3. Year 2:
- New Population: [tex]\( P_2 = P_1 \times 1.13 = 1.13 \times 1.13 \approx 1.2769 \)[/tex]
- Continue to next year as [tex]\( 1.2769 < 2 \)[/tex]
4. Year 3:
- New Population: [tex]\( P_3 = P_2 \times 1.13 = 1.2769 \times 1.13 \approx 1.4439 \)[/tex]
- Continue to next year as [tex]\( 1.4439 < 2 \)[/tex]
5. Year 4:
- New Population: [tex]\( P_4 = P_3 \times 1.13 = 1.4439 \times 1.13 \approx 1.6316 \)[/tex]
- Continue to next year as [tex]\( 1.6316 < 2 \)[/tex]
6. Year 5:
- New Population: [tex]\( P_5 = P_4 \times 1.13 = 1.6316 \times 1.13 \approx 1.8437 \)[/tex]
- Continue to next year as [tex]\( 1.8437 < 2 \)[/tex]
7. Year 6:
- New Population: [tex]\( P_6 = P_5 \times 1.13 = 1.8437 \times 1.13 \approx 2.081952 \)[/tex]
- Stop here, since [tex]\( 2.081952 \)[/tex] exceeds the target of 2.
It takes 6 years for the population to double from 1 to a value exceeding 2, precisely reaching approximately 2.081952.
Therefore, the population of insects will take 6 years to double.
Let's break this down step-by-step:
1. Given Values:
- The growth factor [tex]\( k \)[/tex] is 1.13.
- The initial population [tex]\( P_n \)[/tex] is 1 (for simplicity, we'll start with 1 unit).
- The target population is 2, which is double the initial population.
2. Initialize Variables:
- Start with the initial population, [tex]\( P_n = 1 \)[/tex].
- We need a counter for the number of years, starting at 0.
3. Growth Process:
- Each year, the population grows by multiplying the current population by the growth factor [tex]\( k = 1.13 \)[/tex].
- We keep track of how many years this process takes until the population reaches or exceeds the target.
Let's describe this process step-by-step:
1. Year 0:
- Population: [tex]\( P_0 = 1 \)[/tex]
- Target: 2
2. Year 1:
- New Population: [tex]\( P_1 = P_0 \times 1.13 = 1 \times 1.13 = 1.13 \)[/tex]
- Continue to next year as [tex]\( 1.13 < 2 \)[/tex]
3. Year 2:
- New Population: [tex]\( P_2 = P_1 \times 1.13 = 1.13 \times 1.13 \approx 1.2769 \)[/tex]
- Continue to next year as [tex]\( 1.2769 < 2 \)[/tex]
4. Year 3:
- New Population: [tex]\( P_3 = P_2 \times 1.13 = 1.2769 \times 1.13 \approx 1.4439 \)[/tex]
- Continue to next year as [tex]\( 1.4439 < 2 \)[/tex]
5. Year 4:
- New Population: [tex]\( P_4 = P_3 \times 1.13 = 1.4439 \times 1.13 \approx 1.6316 \)[/tex]
- Continue to next year as [tex]\( 1.6316 < 2 \)[/tex]
6. Year 5:
- New Population: [tex]\( P_5 = P_4 \times 1.13 = 1.6316 \times 1.13 \approx 1.8437 \)[/tex]
- Continue to next year as [tex]\( 1.8437 < 2 \)[/tex]
7. Year 6:
- New Population: [tex]\( P_6 = P_5 \times 1.13 = 1.8437 \times 1.13 \approx 2.081952 \)[/tex]
- Stop here, since [tex]\( 2.081952 \)[/tex] exceeds the target of 2.
It takes 6 years for the population to double from 1 to a value exceeding 2, precisely reaching approximately 2.081952.
Therefore, the population of insects will take 6 years to double.