Water hyacinth is an invasive plant species found in many lakes that typically grows at a rate of [tex]$7 \%$[/tex] per day. As part of a study, a scientist introduces a 150-gram sample of water hyacinth into a testing pool. Which of the following functions gives the amount of water hyacinth in the testing pool [tex]$t$[/tex] weeks after the sample is introduced? (Note: 1 week is 7 days.)

A. [tex]f(t) = 150\left(1 + 0.07^{(1 / 7)}\right)^t[/tex]

B. [tex]g(t) = 150\left(1.07^{(1 / 7)}\right)^t[/tex]

C. [tex]h(t) = 150\left(1 + 0.07^{(7)}\right)^t[/tex]

D. [tex]k(t) = 150\left(1.07^{(7)}\right)^t[/tex]

Answer :

To solve this problem, we need to find the correct mathematical function that describes the growth of water hyacinth given a daily growth rate of 7%. We'll express this growth function in terms of weeks, considering [tex]\( t \)[/tex] is measured in weeks and 1 week equals 7 days.

Here's a step-by-step breakdown of the solution:

1. Understanding the Growth Rate:
- The water hyacinth grows at a rate of 7% per day. In decimal form, this rate is 0.07.

2. Calculate the Weekly Growth Factor:
- Since 1 week is 7 days, we want to determine the growth after 7 days, instead of calculating daily.
- The daily growth factor is [tex]\( 1 + 0.07 = 1.07 \)[/tex].
- To find the weekly growth factor, raise the daily growth factor to the power of 7 (since there are 7 days in a week):
[tex]\[
1.07^7
\][/tex]
- This calculation provides us the total multiplication factor for the entire week.

3. Write the Function:
- The initial amount of water hyacinth is 150 grams.
- The function that gives the amount of water hyacinth [tex]\( t \)[/tex] weeks after introduction, considering weekly growth, is:
[tex]\[
k(t) = 150 \times (1.07^7)^t
\][/tex]

4. Choose the Correct Option:
- Among the given options, we look for the one that matches our derived function:
- Option (D) [tex]\( k(t) = 150(1.07^7)^t \)[/tex] is correct.

Therefore, the function that correctly represents the amount of water hyacinth in the testing pool [tex]\( t \)[/tex] weeks after the sample is introduced is:
[tex]\[
k(t) = 150(1.07^7)^t
\][/tex]

Option D is the correct choice.