A skull cleaning factory cleans arfimal skulls of deer, butfalo and other types of animals using flesh-eating beetles. The factory owner started with only 14 adult beetles. After 37 days, the beetle population grew to 42 adult beetles. Assuming uninhibited exponential growth, how long did it take before the beetle population was 14,000 beetles? It took about $\square$ days. (Do not round until the final answer. Then round to the nearest whole number as noeded.)

Step 1 ：Given the exponential growth formula: \(P = P0 * e^{rt}\), where \(P\) is the final amount, \(P0\) is the initial amount, \(r\) is the rate of growth, and \(t\) is the time.

Step 2 ：We know that \(P0 = 14\), \(P = 42\), and \(t = 37\) days. We can use these values to find the rate of growth (\(r\)).

Step 3 ：Substitute the given values into the formula: \(42 = 14 * e^{37r}\).

Step 4 ：Divide both sides by 14 to get: \(3 = e^{37r}\).

Step 5 ：Take the natural logarithm of both sides to get: \(\ln(3) = 37r\).

Step 6 ：Solve for \(r\) to get: \(r = \frac{\ln(3)}{37}\).

Step 7 ：Now that we have the rate of growth, we can use it to find the time it takes for the population to reach 14,000 beetles.

Step 8 ：Substitute the values into the formula: \(14,000 = 14 * e^{r*t}\).

Step 9 ：Divide both sides by 14 to get: \(1,000 = e^{r*t}\).

Step 10 ：Take the natural logarithm of both sides to get: \(\ln(1,000) = r*t\).

Step 11 ：Substitute \(r = \frac{\ln(3)}{37}\) into the equation to get: \(\ln(1,000) = \frac{\ln(3)}{37} * t\).

Step 12 ：Solve for \(t\) to get: \(t = \frac{37 * \ln(1,000)}{\ln(3)}\).

Step 13 ：Calculate the value to get: \(t \approx 219.9\) days.

Step 14 ：Round to the nearest whole number to get the final answer: \(\boxed{t \approx 220}\) days.