Question A sample of bacteria experienced \( 4 \% \) daily growth compounded continuously for \( d \) days. At the end sample had doubled in size. The doubling-time for this sample of bacteria is \( d \) days. Solve for \( d \) rounding to the nearest whole day. Provide your answer below:

To determine the doubling time for the bacteria, we use the formula for continuous growth, which can be represented by the equation: \[ P(t) = P_0 e^{rt} \] where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( r \) is the growth rate, and \( t \) is the time in days. Since the bacteria doubled in size, we set \( P(t) = 2P_0 \). Given a 4% daily growth rate, we convert this to decimal form \( r = 0.04 \). Substituting our expressions into the formula, we have: \[ 2P_0 = P_0 e^{0.04d} \] Dividing both sides by \( P_0 \) (assuming \( P_0 \neq 0 \)), we simplify to: \[ 2 = e^{0.04d} \] Now, apply the natural logarithm to both sides to solve for \( d \): \[ \ln(2) = 0.04d \] Finally, we isolate \( d \): \[ d = \frac{\ln(2)}{0.04} \] Calculating \( d \): \[ d \approx \frac{0.693147}{0.04} \approx 17.33 \] Rounding to the nearest whole day, we find: \[ d \approx 17 \] So, the doubling-time for this sample of bacteria is \( 17 \) days.