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Use a graphing calculator and the following scenario. The population P of a fish farm in t years is modeled by the equation P(t)=(1700)/(1+9e^(-0.6 t)). To the nearest tenth, how long will it take for the population to reach 900 ? ◻ yr

Use a graphing calculator and the following scenario.\newlineThe population P P of a fish farm in t t years is modeled by the equation P(t)=17001+9e−0.6t P(t)=\frac{1700}{1+9 e^{-0.6 t}} .\newlineTo the nearest tenth, how long will it take for the population to reach 900900 ?\newline□ \square yr

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Q. Use a graphing calculator and the following scenario.\newlineThe population P P of a fish farm in t t years is modeled by the equation P(t)=17001+9e−0.6t P(t)=\frac{1700}{1+9 e^{-0.6 t}} .\newlineTo the nearest tenth, how long will it take for the population to reach 900900 ?\newline□ \square yr
  1. Identify Equation: Identify the equation that models the population of the fish farm over time.\newlineThe given equation is P(t)=17001+9e−0.6tP(t)=\frac{1700}{1+9e^{-0.6 t}}, which is a logistic growth model.
  2. Set Population to 900900: Set the population P(t)P(t) to 900900 to solve for the time tt when the population will reach that number.\newlineSo, we have 900=17001+9e−0.6t900 = \frac{1700}{1+9e^{-0.6 t}}.
  3. Isolate Exponential Part: Isolate the exponential part of the equation by multiplying both sides by the denominator and then dividing by 900900.\newline(\(900)(1+9e−0.6t)=1700(1+9e^{-0.6 t}) = 1700\)\newline1+9e−0.6t=17009001+9e^{-0.6 t} = \frac{1700}{900}\newline1+9e−0.6t=1.88891+9e^{-0.6 t} = 1.8889 (rounded to four decimal places for simplicity).
  4. Subtract to Isolate: Subtract 11 from both sides to isolate the term with the exponential.\newline9e−0.6t=0.88899e^{-0.6 t} = 0.8889
  5. Divide to Solve: Divide both sides by 99 to solve for the exponential term.\newlinee(−0.6t)=0.88899e^{(-0.6 t)} = \frac{0.8889}{9}\newlinee(−0.6t)=0.0988e^{(-0.6 t)} = 0.0988 (rounded to four decimal places for simplicity).
  6. Take Natural Logarithm: Take the natural logarithm of both sides to solve for the exponent.\newline−0.6t=ln⁡(0.0988)-0.6 t = \ln(0.0988)
  7. Divide to Solve for t: Divide both sides by −0.6-0.6 to solve for t.\newlinet=ln⁡(0.0988)−0.6t = \frac{\ln(0.0988)}{-0.6}
  8. Calculate Value of t: Use a calculator to find the value of tt.\newlinet≈ln⁡(0.0988)/(−0.6)t \approx \ln(0.0988)/(-0.6)\newlinet≈2.3148t \approx 2.3148 (rounded to four decimal places for simplicity).
  9. Round to Nearest Tenth: Round the answer to the nearest tenth as the question prompt asks for. t≈2.3t \approx 2.3 years

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