Hello there. To solve this question, we have to remember some properties about exponential growth functions.
Given that the bacteria growing on the water's surface in a non-operational swimming pool from September 20 covers it entirely on September 28, we want to determine when will a quarter of the water's surface be covered.
For this, set the following function that gives you the water surface area covered by the bacteria:
![A(t)=A_0\cdot2^t](https://img.qammunity.org/2023/formulas/mathematics/college/4yygj1br685bguim3nyrjb8a5ghogfuyue.png)
Whereas A0 is the initial area that, for calculation purposes, we consider it as the unit of area, hence A0 = 1 m².
And we know the total area of the pool is given by
![A(8)=A_0\cdot2^8=256A_0=256\text{ m}^2](https://img.qammunity.org/2023/formulas/mathematics/college/hzezomjyhzb1b9moafoflxlfmzdijjxw6f.png)
Now, we need to determine when a quarter of the pool will be covered in bacteria.
In this case, we want that
![A(t)=(A(8))/(4)=(256)/(4)=64](https://img.qammunity.org/2023/formulas/mathematics/college/3qxnr2u8rdyhhikrngcxze4afjnkrrd4uo.png)
In this case, we get that
![\begin{gathered} A(t)=2^t=64 \\ \\ \Rightarrow t=\log_2(64)=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yywg2jgfynvrggvw3l1esa7v7dybjmt1mo.png)
Hence we add it to the initial time observation in days:
![20+6=\text{ September}26\text{ }](https://img.qammunity.org/2023/formulas/mathematics/college/toheg10m9yqmr57h4x6nq0hyav933ng72s.png)
The answer to this question is:
The water's surface will be covered a quarter of the way on September 26