Let's solve the problem step by step. --- **Given:** - The population at time \( t \) hours: \[ P(t) = 100 \times 2^{2t} \] where \( t \) is between 0 and 16. --- ### **a. Find \( P(3) \) and \( P(6) \)** #### **Find \( P(3) \):** \[ P(3) = 100 \times 2^{2 \times 3} = 100 \times 2^{6} = 100 \times 64 = 6400 \] #### **Find \( P(6) \):** \[ P(6) = 100 \times 2^{2 \times 6} = 100 \times 2^{12} = 100 \times 4096 = 409600 \] --- ### **b. Find the time it takes for the E. coli population to number one billion** One billion is \( 1,000,000,000 = 10^9 \). Set \( P(t) = 10^9 \): \[ 100 \times 2^{2t} = 10^9 \] Divide both sides by 100: \[ 2^{2t} = \frac{10^9}{100} = 10^7 \] Take logarithm (base 2) of both sides: \[ 2t = \log_2(10^7) \] Recall: \( \log_2(10^7) = 7 \log_2(10) \) We know \( \log_2(10) \approx 3.3219 \): \[ 2t = 7 \times 3.3219 = 23.2533 \] \[ t = \frac{23.2533}{2} = 11.62665 \] Rounded to the nearest tenth: \[ t \approx 11.6 \] --- ## **Final Answers:** **a.** - \( P(3) = 6400 \) - \( P(6) = 409600 \) **b.** - Time to reach one billion: **11.6 hours** (to the nearest tenth of an hour)

The population doubles every fixed interval, following an exponential growth pattern. To find the time to reach a certain population, we set the growth equation equal to that population and solve for time.