Pink eye is an infection casued by bacteria. After you have been infected, these bacteria double every 40 minutes. How many bacteria would be present at 6 pm if a single bacterium began reproducing at 11am?
$a=1$ and $r=2$ for this geometric sequence and every $40$ min = $2 \over 3$ hrs, so I've come up with the equation
$$\begin{align} t_n&=a(r)^{n-1} \\ t_n&=(2)^{{3\over2}(n-1)} \end{align}$$
$7$ hours passed, so $n=8$
$$\begin{align} t_n&=(2)^{{3\over 2}(8-1)} \\ &=2^{{3\over2}(7)} \\ &=2^{21\over2} \\ &=1448.1546 \end{align}$$
So, there will be $1448$ bacteria.
However, this is wrong. The answer is supposed to be $724$.
Why is my answer wrong? I double checked by creating a chart, and $724$ does not seem reasonable:
$$\begin{array}{|c|c|} \hline \text{Time} & \text{Number of Bacteria} \\\hline 11:00 \text{ am} & 1\\\hline 11:40 \text{ am} & 2\\\hline 12:20 \text{ pm} & 4\\\hline 1:00 \text{ pm} & 8\\\hline 1:40 \text{ pm} & 16\\\hline 2:20 \text{ pm} & 32\\\hline 3:00 \text{ pm} & 64\\\hline 3:40 \text{ pm} & 128\\\hline 4:20 \text{ pm} & 256\\\hline 5:00 \text{ pm} & 512\\\hline 5:40 \text{ pm} & 1024\\\hline 6:00 \text{ pm} & \text{between }1024 \text{ and } 2048 \\\hline 6:20 \text{ pm} & 2048 \\\hline \end{array}$$
Even at 5:40 pm, there should be more than $724$ bacteria. Was there something wrong in my solution or is the correct answer wrong?
Your table is correct until $6:00$ PM. As I read the question the doubling is instantaneous, so at $6:00$ PM there are still $1024$ bacteria. Neither you nor the solution manual should add up all the entries because your table shows the number of bacteria at each time after the doubling. The answer should be $1024$ bacteria at $6:00$PM.