To determine how long it takes for the bacterial culture to lose half its size, follow these steps:
1. Understand the initial condition and desired outcome:
- Initial number of bacteria, [tex]\( B(0) = 9300 \)[/tex]
- We want to find the time [tex]\( t \)[/tex] when [tex]\( B(t) = \frac{9300}{2} = 4650 \)[/tex]
2. Set up the equation using the given model:
[tex]\[
B(t) = 9300 \cdot \left(\frac{1}{64}\right)^t
\][/tex]
We substitute [tex]\( B(t) = 4650 \)[/tex]:
[tex]\[
4650 = 9300 \cdot \left(\frac{1}{64}\right)^t
\][/tex]
3. Solve for [tex]\( t \)[/tex]:
- Divide both sides by 9300:
[tex]\[
\frac{4650}{9300} = \left(\frac{1}{64}\right)^t
\][/tex]
- Simplify the left side:
[tex]\[
\frac{1}{2} = \left(\frac{1}{64}\right)^t
\][/tex]
4. Isolate [tex]\( t \)[/tex]:
- Recognize that we need to solve for [tex]\( t \)[/tex] in the equation: [tex]\( \left(\frac{1}{64}\right)^t = \frac{1}{2} \)[/tex].
5. Utilize logarithms (optional for understanding, but not necessary as we have the calculated result):
- If using logarithms, it could be something like:
[tex]\[
t = \frac{\ln(\frac{1}{2})}{\ln(\frac{1}{64})}
\][/tex]
6. State the solution:
- From the calculations, we find that the time [tex]\( t \)[/tex] when the bacteria count is halved is approximately:
[tex]\[
t \approx 0.17 \text{ seconds}
\][/tex]
So, the bacterial culture loses [tex]\(\frac{1}{2}\)[/tex] of its size every 0.17 seconds.
