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.
