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Bacteria begins to grow on the water's surface in a non-operational swimming pool on September 20. The bacteria grows and covers the water'ssurface in such a way that the area covered with bacteria doubles every day. If It continues to grow in this way, the water's surface will beentirely covered with bacteria on September 28.When will a quarter of the water's surface be covered?

Bacteria begins to grow on the water's surface in a non-operational swimming pool-example-1

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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

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

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

In this case, we get that


\begin{gathered} A(t)=2^t=64 \\ \\ \Rightarrow t=\log_2(64)=6 \end{gathered}

Hence we add it to the initial time observation in days:


20+6=\text{ September}26\text{ }

The answer to this question is:

The water's surface will be covered a quarter of the way on September 26

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