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 can set up a table to track the population growth over time.
Starting with 60 hydra in the initial population, we can use the doubling rate of every two days to calculate the population for each subsequent period.
| Period | Population |
|--------|------------|
| 0 | 60 |
| 1 | 120 |
| 2 | 240 |
| 3 | 480 |
| 4 | 960 |
| 5 | 1920 |
| 6 | 3840 |
| 7 | 7680 |
Looking at the table, we can see that after the 7th two-day period, the population exceeds 5000. Therefore, the correct answer is "after 7 two-day periods."
To model this situation mathematically, let's use the function 7 = a * b^(x), where a is the initial population (60), b is the growth factor (2 in this case), and x represents the number of two-day periods.
In this case, we want to find the value of x when the population exceeds 5000. Let's substitute the known values into our equation:
5000 = 60 * 2^(x)
To solve for x, we need to isolate the variable. We can start by dividing both sides of the equation by 60:
5000/60 = 2^(x)
83.33 = 2^(x)
Next, we can take the logarithm of both sides to eliminate the exponent:
log(83.33) = log(2^(x))
Using the logarithmic property, we can bring the exponent down:
log(83.33) = x * log(2)
Now, we can solve for x by dividing both sides of the equation by log(2):
x = log(83.33) / log(2)
Using a calculator, we find that x is approximately 6.092, which means after approximately 6.092 two-day periods, the population will exceed 5000.
Therefore, the correct answer is that there will be more than 5000 hydra after approximately 7 two-day periods.
If t is the time in days after starting, and No is the initial number,
N(t) = No* 2^(t/2)
5000 = 60*2^(t/2)
2^(t/2) = 83.3
t/2 = log83.3/log2 = 6.38
t = 12.76 days