Ellis Bird
03/14/2024 · Junior High School
A population of bacteria is growing according to the equation \( P(t)=1700 e^{0.08 t} \) Use a graphing calculator to estimate when the population will exceed 2476
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Step-by-step Solution
To find when the population \( P(t) \) exceeds 2476, we need to solve the inequality:
\[
1700 e^{0.08 t} > 2476
\]
First, we can divide both sides by 1700:
\[
e^{0.08 t} > \frac{2476}{1700}
\]
Calculating the right side:
\[
\frac{2476}{1700} \approx 1.45
\]
Now we have:
\[
e^{0.08 t} > 1.45
\]
Next, we take the natural logarithm of both sides:
\[
0.08 t > \ln(1.45)
\]
Calculating \( \ln(1.45) \):
\[
\ln(1.45) \approx 0.370
\]
Now we can solve for \( t \):
\[
t > \frac{0.370}{0.08}
\]
Calculating the right side:
\[
t > 4.625
\]
Thus, the population will exceed 2476 at approximately \( t \approx 4.625 \).
To find the exact time, we can round this to the nearest whole number or decimal place depending on the context. If we round to the nearest whole number, it would be \( t = 5 \).
To confirm, we can check the population at \( t = 4 \) and \( t = 5 \):
1. For \( t = 4 \):
\[
P(4) = 1700 e^{0.08 \times 4} \approx 1700 e^{0.32} \approx 1700 \times 1.377 \approx 2341.0
\]
2. For \( t = 5 \):
\[
P(5) = 1700 e^{0.08 \times 5} \approx 1700 e^{0.4} \approx 1700 \times 1.491 \approx 2534.0
\]
Since \( P(4) \approx 2341.0 < 2476 \) and \( P(5) \approx 2534.0 > 2476 \), we confirm that the population exceeds 2476 between \( t = 4 \) and \( t = 5 \).
Thus, the population will exceed 2476 at approximately \( t \approx 4.625 \) or around \( t = 5 \).
Quick Answer
The population will exceed 2476 at approximately \( t \approx 4.625 \) or around \( t = 5 \).
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