\begin{tabular}{l} Evaluating and Solving Exponential Functions \\ \hline Since 1991, the number of fish in Lake Beckett has been decreasing at a rate of \( 1.7 \% \) per year. In 1991, the \\ population of fish was estimated to be 101 million. Use this information to answer the following: \\ \hline a) Write the exponential function \( P(t) \) for this scenario where \( P(t) \) is the fish population in millions \( t \) \\ years after 1991. \\ Exponential Function: \( P(t)= \) \\ b) Determine the number of fish in Lake Beckett in 1996. Round your answer to two decimal places. \\ The population of fish in Lake Beckett in 1996 will be \\ \hline \( \begin{array}{l}\text { c) Determine in what year the population of fish will be half the amount it was in } 1991 .\end{array} \) Round your answer \\ to the nearest year. \\ The population of fish in Lake Beckett will be half of what it was in 1991 in the year \\ \hline\end{tabular}

To solve the problem, we will go through each part step by step. ### a) Write the exponential function \( P(t) \) The general form of an exponential decay function is given by: \[ P(t) = P_0 \cdot (1 - r)^t \] where: - \( P_0 \) is the initial population, - \( r \) is the decay rate (as a decimal), - \( t \) is the time in years after the initial year. In this case: - The initial population \( P_0 = 101 \) million, - The decay rate \( r = 1.7\% = 0.017 \). Thus, the function becomes: \[ P(t) = 101 \cdot (1 - 0.017)^t \] This simplifies to: \[ P(t) = 101 \cdot (0.983)^t \] ### b) Determine the number of fish in Lake Beckett in 1996 To find the population in 1996, we need to calculate \( P(t) \) for \( t = 1996 - 1991 = 5 \): \[ P(5) = 101 \cdot (0.983)^5 \] Calculating \( (0.983)^5 \): \[ (0.983)^5 \approx 0.8681 \] Now, substituting back into the equation: \[ P(5) \approx 101 \cdot 0.8681 \approx 87.66 \] Thus, the population of fish in Lake Beckett in 1996 will be approximately **87.66 million**. ### c) Determine in what year the population of fish will be half the amount it was in 1991 To find when the population will be half of the initial population, we set up the equation: \[ P(t) = \frac{101}{2} \] Substituting the function: \[ 101 \cdot (0.983)^t = 50.5 \] Dividing both sides by 101: \[ (0.983)^t = \frac{50.5}{101} \approx 0.500 \] Now, we take the natural logarithm of both sides: \[ \ln((0.983)^t) = \ln(0.500) \] Using the property of logarithms: \[ t \cdot \ln(0.983) = \ln(0.500) \] Now, solving for \( t \): \[ t = \frac{\ln(0.500)}{\ln(0.983)} \] Calculating the logarithms: \[ \ln(0.500) \approx -0.6931 \] \[ \ln(0.983) \approx -0.0170 \] Now substituting these values: \[ t \approx \frac{-0.6931}{-0.0170} \approx 40.74 \] Rounding to the nearest year, we find \( t \approx 41 \). Since \( t \) is the number of years after 1991, we add this to 1991: \[ 1991 + 41 = 2032 \] Thus, the population of fish in Lake Beckett will be half of what it was in 1991 in the year **2032**. ### Summary of Answers: - a) Exponential Function: \( P(t) = 101 \cdot (0.983)^t \) - b) The population of fish in Lake Beckett in 1996 will be approximately **87.66 million**. - c) The population of fish in Lake Beckett will be half of what it was in 1991 in the year **2032**.