A colony of bacteria grows exponentially at a rate of \( 2 / 7 \) every day. The colony begins with 2000 cells. How many cells are there after three weeks? Remember not to round until you get your final value. - Leave values as fractions don't use decimals which would require rounding. Round your final answer to the nearest whole number of cells. There will be cells after three weeks.

To find the number of cells after three weeks, we can use the formula for exponential growth: \[ N(t) = N_0 \cdot e^{kt} \] However, in this case, we'll use the simpler model for exponential growth in discrete time: \[ N(t) = N_0 \cdot \left(1 + r\right)^t \] where \(N_0\) is the initial quantity, \(r\) is the growth rate, and \(t\) is time in days. Here, we know: - \(N_0 = 2000\) - \(r = \frac{2}{7}\) - \(t = 21\) days (3 weeks) Now, plug in the values: \[ N(21) = 2000 \cdot \left(1 + \frac{2}{7}\right)^{21} \] First, calculate \(1 + \frac{2}{7} = \frac{9}{7}\). Then, raise \(\frac{9}{7}\) to the power of \(21\): \[ N(21) = 2000 \cdot \left(\frac{9}{7}\right)^{21} \] Calculating \(\left(\frac{9}{7}\right)^{21}\) gives: \[ \left(\frac{9^{21}}{7^{21}}\right) \] Now, we continue: \[ N(21) = 2000 \cdot \frac{9^{21}}{7^{21}} = \frac{2000 \cdot 9^{21}}{7^{21}} \] Now, let's calculate the complete fraction: \[ N(21) = \frac{2000 \cdot 9^{21}}{7^{21}} \] To find the final number of cells, you would then compute this fraction and round it to the nearest whole number. However, since we are not rounding until the end, let's consider it still as a fraction: After calculating the values: \[ N(21) ≈ 81718190581742248568 \text{ (this value is } \frac{2000 \cdot 9^{21}}{7^{21}} \text{ without performing the rounding)} \] After rounding this final answer, we find: There will be **81718190581742248600** cells after three weeks.