Hydra are small freshwater animals. They can double in number every two days in a laboratory tank. Suppose one tank has an initial population of 60 hydra. When will there be more than 5000 hydra? How can a table help you identify a pattern? What function models the situation (remember that in the function 7=a*b(to the power of x)? I believe the answer is "after 7 2-day periods", but I do not know how to set it up.
To determine when there will be more than 5000 hydra in the tank, we need to identify the pattern of hydra growth and set up a table to track the population over time.
Let's start by understanding the pattern of hydra population growth. We are given that the hydra population doubles every two days. This implies exponential growth, where the population size, P, can be modeled using the formula:
P = P₀ * 2^(x/2)
Where P₀ is the initial population size and x is the number of 2-day periods.
Now, let's set up a table to track the population over time:
| Time Period | Number of Hydra |
|-------------|----------------|
| 0 | 60 |
| 2 | ? |
| 4 | ? |
| 6 | ? |
| 8 | ? |
| ... | ... |
We already know that the initial population (at time period 0) is 60 hydra. To find the number of hydra at future time periods, we'll follow the exponential growth formula.
For each time period, we'll use the formula P = 60 * 2^(x/2), where x is the number of 2-day periods. Let's calculate the population for each time period:
| Time Period | Number of Hydra |
|-------------|----------------|
| 0 | 60 |
| 2 | 60 * 2^1 |
| 4 | 60 * 2^2 |
| 6 | 60 * 2^3 |
| 8 | 60 * 2^4 |
| ... | ... |
To find the number of hydra at each time period, we need to calculate 2^(x/2), where x is the time period. For example, at time period 2, we calculate 2^(2/2) = 2^1 = 2. So, the number of hydra will be 60 * 2^1 = 120.
Continuing this process, we can fill in the table:
| Time Period | Number of Hydra |
|-------------|---------------------------|
| 0 | 60 |
| 2 | 60 * 2^1 = 120 |
| 4 | 60 * 2^2 = 240 |
| 6 | 60 * 2^3 = 480 |
| 8 | 60 * 2^4 = 960 |
| 10 | 60 * 2^5 = 1920 |
| 12 | 60 * 2^6 = 3840 |
| 14 | 60 * 2^7 = 7680 |
From the table, we can observe that at time period 14, the number of hydra exceeds 5000. Therefore, the answer is that there will be more than 5000 hydra after 14 2-day periods.
In terms of the function that models the situation, we can use the formula P = 60 * 2^(x/2), where P is the population size and x is the number of 2-day periods.