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

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:

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