123. Sheep population. A herd of sheep doubles in size every three years. There are now 1500 sheep in the herd. a. Find an exponential function of the form \( P=P_{0} e^{r t} \) to model the herd's growth. b. How many sheep will be in the herd seven years from now (round your answer)? c. In how many years will the herd have 15,000 sheep?

To find the exponential function, we start with the given parameters. Since the herd doubles every three years, we can find the growth rate \( r \). a. The initial population \( P_0 = 1500 \), and after 3 years, the population will double to \( 3000 \). We can set up the equation: \[ 3000 = 1500 e^{3r} \] Dividing both sides by 1500 gives: \[ 2 = e^{3r} \] Taking the natural logarithm of both sides: \[ \ln(2) = 3r \] \[ r = \frac{\ln(2)}{3} \] Thus, the exponential function can be formulated as: \[ P(t) = 1500 e^{\frac{\ln(2)}{3}t} \] b. To find the sheep population in 7 years, we evaluate: \[ P(7) = 1500 e^{\frac{\ln(2)}{3} \times 7} \approx 1500 e^{\ln(2) \times \frac{7}{3}} \] Calculating \( e^{\ln(2) \times \frac{7}{3}} \) simplifies to \( 2^{\frac{7}{3}} \approx 10.0794 \). Therefore, \[ P(7) \approx 1500 \times 10.0794 \approx 15119 \] Rounding gives approximately **15,119 sheep**. c. To find when the herd will reach 15,000 sheep, we solve for \( t \): \[ 15000 = 1500 e^{\frac{\ln(2)}{3} t} \] Dividing by 1500 yields: \[ 10 = e^{\frac{\ln(2)}{3} t} \] Taking the natural logarithm: \[ \ln(10) = \frac{\ln(2)}{3} t \] Solving for \( t \) gives: \[ t = \frac{3 \ln(10)}{\ln(2)} \approx \frac{3 \times 2.3026}{0.6931} \approx 9.97 \] Therefore, **in about 10 years**, the herd will have reached 15,000 sheep.