If the doubling time of a bacterial population is 3 hours, then its average specific growth rate during this period is ________ h\(^{-1}\). (Round off to two decimal places)
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Solution and Explanation
\( \ln(2) \) is the natural logarithm of 2, and
\( t_d \) is the doubling time in hours.
Given that \( t_d = 3 \) hours, we can substitute this value into the equation: \[ \mu = \frac{\ln(2)}{3} \] Now, calculate the value: \[ \mu = \frac{0.6931}{3} \approx 0.231 { h}^{-1}. \] Thus, the average specific growth rate is \( \boxed{0.23} \) h\(^{-1}\).
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