The population of bacteria in a container follows the law of uninhibited growth. At the beginning, there were 9 grams of bacteria. After 11 hours, there were 17.41 grams of bacteria. Answer the following question. After $\square$ hours, there would be 27 grams of bacteria. Round your answer to one decimal place.

Step 1 ：Given that \(P_0 = 9\) grams, \(P = 17.41\) grams, and \(t = 11\) hours, we substitute these values into the equation \(P = P_0 \times e^{kt}\) to get \(17.41 = 9 \times e^{11k}\)

Step 2 ：Divide both sides by 9 to get \(1.934 = e^{11k}\)

Step 3 ：Take the natural logarithm of both sides to get \(\ln(1.934) = 11k\) or \(0.660 = 11k\)

Step 4 ：Divide by 11 to solve for \(k\) to get \(k = \frac{0.660}{11} = 0.060\)

Step 5 ：Now that we have the growth rate, we can use it to find the time it would take for the bacteria to reach 27 grams. Substitute \(P = 27\), \(P_0 = 9\), and \(k = 0.060\) into the equation \(P = P_0 \times e^{kt}\) to get \(27 = 9 \times e^{0.060t}\)

Step 6 ：Divide both sides by 9 to get \(3 = e^{0.060t}\)

Step 7 ：Take the natural logarithm of both sides to get \(\ln(3) = 0.060t\) or \(1.099 = 0.060t\)

Step 8 ：Divide by 0.060 to solve for \(t\) to get \(t = \frac{1.099}{0.060} = 18.3\) hours

Step 9 ：So, after approximately 18.3 hours, there would be 27 grams of bacteria. Therefore, \(\boxed{t = 18.3}\) hours