Question:
A population shows exponential growth of the form \( N_t = N_0 e^{rt} \) where \( N_t \) is the population at time \( t \), \( N_0 \) is the initial population size, and \( r \) is the rate of increase. If \( r = 0.1 \), then the doubling time for this population is \(\underline{\hspace{1cm}}\) (Round off to two decimal places.)
A population shows exponential growth of the form \( N_t = N_0 e^{rt} \) where \( N_t \) is the population at time \( t \), \( N_0 \) is the initial population size, and \( r \) is the rate of increase. If \( r = 0.1 \), then the doubling time for this population is \(\underline{\hspace{1cm}}\) (Round off to two decimal places.)
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The doubling time for an exponentially growing population is calculated as \( T_d = \frac{\ln(2)}{r} \).
Updated On: Dec 24, 2025
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Correct Answer: 6.8
Solution and Explanation
The doubling time \( T_d \) is the time it takes for the population to double, and it can be found using the formula:
\[
T_d = \frac{\ln(2)}{r}.
\]
Given \( r = 0.1 \), we calculate:
\[
T_d = \frac{\ln(2)}{0.1} \approx \frac{0.693}{0.1} = 6.93.
\]
Thus, the doubling time for this population is \( \boxed{6.93} \) years.
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