$$\text{Growth of a Culture of Bacteria}$$ $$\begin{array} {|c|c|} \hline & \text{Number of bacteria per } \\ \text{Day} & \text{milliliter at the end of day} \\ \hline 1 & 3.5 \times 10^3 \\ \hline 2 & 7.0 \times 10^3 \\ \hline 3 & 1.4 \times 10^4 \\ \hline \end{array}$$

A culture of bacteria is growing at an exponential rate, as shown in the table above. At this rate, on which day would the number of bacteria per milliliter reach $1.12 \times 10^5$?

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With a bit of mental math or utilizing the calculator, we can extend the table and inspect when the bacteria will reach the desired density.

$$\begin{array} {|c|c|} \hline & \text{Number of bacteria per } \\ \text{Day} & \text{milliliter at the end of day} \\ \hline 1 & 3.5 \times 10^3 \\ \hline 2 & 7.0 \times 10^3 \\ \hline 3 & 1.4 \times 10^4 \\ \hline 4 & 2.8 \times 10^4 \\ \hline 5 & 5.6 \times 10^4 \\ \hline \colorbox{aqua}{$6$} & \colorbox{aqua}{$1.12 \times 10^5$} \\ \hline \end{array}$$

If we were to model the situation, we can start with the initial value of $3.5 \times 10^3$. Through inspection, we see that the bacteria is doubling every day. We can use the following equation:

$$y = (3.5\times 10^3) (2)^x$$ $$\text{where } x= \text{ number of days after day 1}$$

Plugging in our desired final value:

$$1.12 \times 10^5 = (3.5\times 10^3) (2)^x$$ $$\frac{1.12 \times 10^5}{3.5 \times 10^3} = 2^x$$ $$32=2^x$$ $$2^5=2^x$$ $$x=5 \text{ days after day 1}$$ $$\boxed{\text{Day 6}}$$