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Sagot :
Let's solve the given equation step-by-step to find out how long it will take for the population of the fish farm to reach 900.
We start with the equation provided for the population [tex]\( P \)[/tex] at time [tex]\( t \)[/tex]:
[tex]\[ P(t) = \frac{1800}{1 + 9e^{-0.7t}} \][/tex]
We need to determine the time [tex]\( t \)[/tex] when the population [tex]\( P(t) \)[/tex] equals 900. So, we set up the equation:
[tex]\[ 900 = \frac{1800}{1 + 9e^{-0.7t}} \][/tex]
1. Set up the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ 900 = \frac{1800}{1 + 9e^{-0.7t}} \][/tex]
2. Isolate the fraction:
Multiply both sides by [tex]\( 1 + 9e^{-0.7t} \)[/tex] to get rid of the fraction:
[tex]\[ 900(1 + 9e^{-0.7t}) = 1800 \][/tex]
3. Simplify the equation:
[tex]\[ 900 + 900 \cdot 9e^{-0.7t} = 1800 \][/tex]
[tex]\[ 900 + 8100e^{-0.7t} = 1800 \][/tex]
4. Isolate the exponential term:
Subtract 900 from both sides:
[tex]\[ 8100e^{-0.7t} = 900 \][/tex]
5. Solve for the exponential term:
Divide both sides by 8100:
[tex]\[ e^{-0.7t} = \frac{900}{8100} \][/tex]
[tex]\[ e^{-0.7t} = \frac{1}{9} \][/tex]
6. Take the natural logarithm of both sides:
To solve for [tex]\( t \)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[ \ln(e^{-0.7t}) = \ln\left(\frac{1}{9}\right) \][/tex]
Using the properties of logarithms, we know that [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ -0.7t = \ln\left(\frac{1}{9}\right) \][/tex]
7. Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{1}{9}\right)}{-0.7} \][/tex]
8. Calculate the values:
The natural logarithm of [tex]\( \frac{1}{9} \)[/tex] is approximately -2.1972. Plugging this into our equation:
[tex]\[ t \approx \frac{-2.1972}{-0.7} \approx 3.1389 \][/tex]
9. Round to the nearest tenth:
Therefore, [tex]\( t \)[/tex] rounded to the nearest tenth is approximately 3.1.
So, it will take approximately 3.1 years for the population of the fish farm to reach 900.
We start with the equation provided for the population [tex]\( P \)[/tex] at time [tex]\( t \)[/tex]:
[tex]\[ P(t) = \frac{1800}{1 + 9e^{-0.7t}} \][/tex]
We need to determine the time [tex]\( t \)[/tex] when the population [tex]\( P(t) \)[/tex] equals 900. So, we set up the equation:
[tex]\[ 900 = \frac{1800}{1 + 9e^{-0.7t}} \][/tex]
1. Set up the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ 900 = \frac{1800}{1 + 9e^{-0.7t}} \][/tex]
2. Isolate the fraction:
Multiply both sides by [tex]\( 1 + 9e^{-0.7t} \)[/tex] to get rid of the fraction:
[tex]\[ 900(1 + 9e^{-0.7t}) = 1800 \][/tex]
3. Simplify the equation:
[tex]\[ 900 + 900 \cdot 9e^{-0.7t} = 1800 \][/tex]
[tex]\[ 900 + 8100e^{-0.7t} = 1800 \][/tex]
4. Isolate the exponential term:
Subtract 900 from both sides:
[tex]\[ 8100e^{-0.7t} = 900 \][/tex]
5. Solve for the exponential term:
Divide both sides by 8100:
[tex]\[ e^{-0.7t} = \frac{900}{8100} \][/tex]
[tex]\[ e^{-0.7t} = \frac{1}{9} \][/tex]
6. Take the natural logarithm of both sides:
To solve for [tex]\( t \)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[ \ln(e^{-0.7t}) = \ln\left(\frac{1}{9}\right) \][/tex]
Using the properties of logarithms, we know that [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ -0.7t = \ln\left(\frac{1}{9}\right) \][/tex]
7. Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{1}{9}\right)}{-0.7} \][/tex]
8. Calculate the values:
The natural logarithm of [tex]\( \frac{1}{9} \)[/tex] is approximately -2.1972. Plugging this into our equation:
[tex]\[ t \approx \frac{-2.1972}{-0.7} \approx 3.1389 \][/tex]
9. Round to the nearest tenth:
Therefore, [tex]\( t \)[/tex] rounded to the nearest tenth is approximately 3.1.
So, it will take approximately 3.1 years for the population of the fish farm to reach 900.
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