A bacteria culture starts with 460 bacteria and grows at a rate proportional to its size. After 2 hours there will be 920 bacteria. (a) Express the population $P$ after $t$ hours as a function of $t$. \[ P(t)= \] (b) What will be the population after 6 hours? \[ 3680 \] Q bacteria (c) How long will it take for the population to reach 2790? Give your answer accurate to at least 2 decimal places. hours

Step 1 ：Given that the initial population \(P_0\) is 460 and after 2 hours the population is 920, we can substitute these values into the equation to find the growth rate \(k\): \[920 = 460 * e^{2k}\]

Step 2 ：Divide both sides by 460: \[2 = e^{2k}\]

Step 3 ：Take the natural logarithm of both sides: \[\ln(2) = 2k\]

Step 4 ：Solve for \(k\): \[k = \frac{\ln(2)}{2}\]

Step 5 ：Substitute \(k\) back into the original equation: \[P(t) = 460 * e^{t * \frac{\ln(2)}{2}}\]

Step 6 ：To find the population after 6 hours, substitute \(t = 6\) into the equation: \[P(6) = 460 * e^{6 * \frac{\ln(2)}{2}} = 3680\]

Step 7 ：To find the time it takes for the population to reach 2790, set \(P(t) = 2790\) and solve for \(t\): \[2790 = 460 * e^{t * \frac{\ln(2)}{2}}\]

Step 8 ：Divide both sides by 460: \[6.065217391 = e^{t * \frac{\ln(2)}{2}}\]

Step 9 ：Take the natural logarithm of both sides: \[\ln(6.065217391) = t * \frac{\ln(2)}{2}\]

Step 10 ：Solve for \(t\): \[t = \frac{2 * \ln(6.065217391)}{\ln(2)}\]

Step 11 ：\(\boxed{t \approx 5.26 \text{ hours}}\)