Sign In
Make a table showing the number of hydra and days by using the information. Then, write a function representing the population growth.
After 12 days
We will follow three steps to solve the question.
Hydra can double in number every two days in a laboratory tank. The tank has an initial population of 60 hydra. Using this information, we will make a table showing the number of hydra depending on days. Let x represent the number of days and y represent the number of hydra.
There is a common difference (2) between x-values and a common ratio ( 2) between y-values.
Since there is a common ratio between y-values, the table represents an exponential function. Any exponential function has the following form. y= a* b^x Here, a is the initial value and b is the constant multiplier. Now we can partially write the function since 60 is the inital amount and 2 is the constant multiplier. y= a* b^x ⇔ y= 60 * 2^x This function does not represent the situation exactly because when we substitute 1 for x, it will double the number. That is, the population doubles itself every day. However, we want that the population doubles itself every two days. To do so, we need to divide x by 2. y=60 * 2^x ⇒ y=60 * 2^(x2) As a result, y=60 * 2^(x2) models the situation.
To find when there will be more than 5000 hydra, we will make a table of values for the function. We will use the values of x greater than 6 since we know that the population is 480, less than 5000. We will finish the table when we find a value more than 5000. Let's do it.
x | 60 * 2^(x2) | y=60 * 2^(x2) |
---|---|---|
8 | 60 * 2^(82) | 960 |
10 | 60 * 2^(102) | 1920 |
12 | 60 * 2^(122) | 3840 |
14 | 60 * 2^(142) | 7680 |
From the table, we see that 5000 is between 3840 and 7680, which are the population after 12 and 14 days respectively. Therefore, the hydra population will be more than 5000 after 12 days.