Question
The count in a bacteria culture was 200 after 20 minutes and 1200 after 35 minutes. Assuming the count grows exponentially,
What was the initial size of the culture?
What was the initial size of the culture?
Answers
Answered by
GPT-4o mini
To find the initial size of the bacterial culture, let's denote the initial size by \( N_0 \) and the growth rate by \( r \). The formula for exponential growth is given by:
\[
N(t) = N_0 e^{rt}
\]
where \( N(t) \) is the population at time \( t \), \( N_0 \) is the initial population, \( r \) is the growth rate, and \( t \) is time.
We know the following:
1. At \( t = 20 \) minutes, \( N(20) = 200 \):
\[
200 = N_0 e^{20r}
\]
2. At \( t = 35 \) minutes, \( N(35) = 1200 \):
\[
1200 = N_0 e^{35r}
\]
We can set up a system of equations based on the two observations:
1. \( 200 = N_0 e^{20r} \) (Equation 1)
2. \( 1200 = N_0 e^{35r} \) (Equation 2)
To eliminate \( N_0 \), we can divide Equation 2 by Equation 1:
\[
\frac{1200}{200} = \frac{N_0 e^{35r}}{N_0 e^{20r}}
\]
This simplifies to:
\[
6 = e^{35r - 20r}
\]
\[
6 = e^{15r}
\]
Taking the natural logarithm of both sides gives:
\[
\ln(6) = 15r
\]
\[
r = \frac{\ln(6)}{15}
\]
Now, we can substitute \( r \) back into Equation 1 to find \( N_0 \):
\[
200 = N_0 e^{20 \cdot \frac{\ln(6)}{15}}
\]
\[
200 = N_0 e^{\frac{20 \ln(6)}{15}} = N_0 e^{\frac{4 \ln(6)}{3}}
\]
Rearranging this equation to isolate \( N_0 \):
\[
N_0 = 200 e^{-\frac{4 \ln(6)}{3}}
\]
\[
N_0 = 200 \cdot 6^{-\frac{4}{3}}
\]
Calculating \( 6^{-\frac{4}{3}} \):
\[
6^{-\frac{4}{3}} = \frac{1}{6^{\frac{4}{3}}} = \frac{1}{(6^{4})^{\frac{1}{3}}} = \frac{1}{1296^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{1296}}
\]
Since \( 6^4 = 1296 \),
\[
\sqrt[3]{1296} \approx 10.0794
\]
Thus,
\[
N_0 = \frac{200}{10.0794} \approx 19.8
\]
Therefore, the initial size of the culture was approximately \( 19.8 \), or rounding appropriately, we can say the initial size was about **20 bacteria**.
\[
N(t) = N_0 e^{rt}
\]
where \( N(t) \) is the population at time \( t \), \( N_0 \) is the initial population, \( r \) is the growth rate, and \( t \) is time.
We know the following:
1. At \( t = 20 \) minutes, \( N(20) = 200 \):
\[
200 = N_0 e^{20r}
\]
2. At \( t = 35 \) minutes, \( N(35) = 1200 \):
\[
1200 = N_0 e^{35r}
\]
We can set up a system of equations based on the two observations:
1. \( 200 = N_0 e^{20r} \) (Equation 1)
2. \( 1200 = N_0 e^{35r} \) (Equation 2)
To eliminate \( N_0 \), we can divide Equation 2 by Equation 1:
\[
\frac{1200}{200} = \frac{N_0 e^{35r}}{N_0 e^{20r}}
\]
This simplifies to:
\[
6 = e^{35r - 20r}
\]
\[
6 = e^{15r}
\]
Taking the natural logarithm of both sides gives:
\[
\ln(6) = 15r
\]
\[
r = \frac{\ln(6)}{15}
\]
Now, we can substitute \( r \) back into Equation 1 to find \( N_0 \):
\[
200 = N_0 e^{20 \cdot \frac{\ln(6)}{15}}
\]
\[
200 = N_0 e^{\frac{20 \ln(6)}{15}} = N_0 e^{\frac{4 \ln(6)}{3}}
\]
Rearranging this equation to isolate \( N_0 \):
\[
N_0 = 200 e^{-\frac{4 \ln(6)}{3}}
\]
\[
N_0 = 200 \cdot 6^{-\frac{4}{3}}
\]
Calculating \( 6^{-\frac{4}{3}} \):
\[
6^{-\frac{4}{3}} = \frac{1}{6^{\frac{4}{3}}} = \frac{1}{(6^{4})^{\frac{1}{3}}} = \frac{1}{1296^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{1296}}
\]
Since \( 6^4 = 1296 \),
\[
\sqrt[3]{1296} \approx 10.0794
\]
Thus,
\[
N_0 = \frac{200}{10.0794} \approx 19.8
\]
Therefore, the initial size of the culture was approximately \( 19.8 \), or rounding appropriately, we can say the initial size was about **20 bacteria**.