Question
Question 4 (1 point)
The population of bacteria in a petri dish doubles (b) every 24 hours (n). The population of the bacteria started at 500 organisms (a).
How long will it take (t) for the population of the bacteria to reach 800 (y)?
Round your answer to the nearest tenth of an hour.
Question 4 options:
16.3 hours
17.4 hours
13.8 hours
14.1 hours
The population of bacteria in a petri dish doubles (b) every 24 hours (n). The population of the bacteria started at 500 organisms (a).
How long will it take (t) for the population of the bacteria to reach 800 (y)?
Round your answer to the nearest tenth of an hour.
Question 4 options:
16.3 hours
17.4 hours
13.8 hours
14.1 hours
Answers
GPT-4o mini
To find the time it takes for the population of bacteria to grow from an initial amount to a desired amount, we can use the exponential growth formula:
\[
P(t) = a \cdot b^{(t/n)}
\]
Where:
- \(P(t)\) is the population at time \(t\)
- \(a\) is the initial population
- \(b\) is the growth factor (2 for doubling)
- \(t\) is the time in hours
- \(n\) is the time it takes to double the population (24 hours in this case)
In this scenario:
- \(P(t) = 800\)
- \(a = 500\)
- \(b = 2\)
- \(n = 24\)
Now, we can set up the equation:
\[
800 = 500 \cdot 2^{(t/24)}
\]
To isolate \(2^{(t/24)}\), divide both sides by 500:
\[
\frac{800}{500} = 2^{(t/24)}
\]
This simplifies to:
\[
1.6 = 2^{(t/24)}
\]
To solve for \(t\), we take the logarithm (base 2) of both sides:
\[
\log_2(1.6) = \frac{t}{24}
\]
Now, we can solve for \(t\):
\[
t = 24 \cdot \log_2(1.6)
\]
We can calculate \(\log_2(1.6)\) using the change of base formula:
\[
\log_2(1.6) = \frac{\log_{10}(1.6)}{\log_{10}(2)}
\]
Using a calculator:
- \(\log_{10}(1.6) \approx 0.2041\)
- \(\log_{10}(2) \approx 0.3010\)
Therefore:
\[
\log_2(1.6) \approx \frac{0.2041}{0.3010} \approx 0.678
\]
Now substituting this back into the equation for \(t\):
\[
t \approx 24 \cdot 0.678 \approx 16.3 \text{ hours}
\]
Thus, the time it takes for the population of bacteria to reach 800 organisms is approximately **16.3 hours**.
**Final answer: 16.3 hours**.
\[
P(t) = a \cdot b^{(t/n)}
\]
Where:
- \(P(t)\) is the population at time \(t\)
- \(a\) is the initial population
- \(b\) is the growth factor (2 for doubling)
- \(t\) is the time in hours
- \(n\) is the time it takes to double the population (24 hours in this case)
In this scenario:
- \(P(t) = 800\)
- \(a = 500\)
- \(b = 2\)
- \(n = 24\)
Now, we can set up the equation:
\[
800 = 500 \cdot 2^{(t/24)}
\]
To isolate \(2^{(t/24)}\), divide both sides by 500:
\[
\frac{800}{500} = 2^{(t/24)}
\]
This simplifies to:
\[
1.6 = 2^{(t/24)}
\]
To solve for \(t\), we take the logarithm (base 2) of both sides:
\[
\log_2(1.6) = \frac{t}{24}
\]
Now, we can solve for \(t\):
\[
t = 24 \cdot \log_2(1.6)
\]
We can calculate \(\log_2(1.6)\) using the change of base formula:
\[
\log_2(1.6) = \frac{\log_{10}(1.6)}{\log_{10}(2)}
\]
Using a calculator:
- \(\log_{10}(1.6) \approx 0.2041\)
- \(\log_{10}(2) \approx 0.3010\)
Therefore:
\[
\log_2(1.6) \approx \frac{0.2041}{0.3010} \approx 0.678
\]
Now substituting this back into the equation for \(t\):
\[
t \approx 24 \cdot 0.678 \approx 16.3 \text{ hours}
\]
Thus, the time it takes for the population of bacteria to reach 800 organisms is approximately **16.3 hours**.
**Final answer: 16.3 hours**.