Question

the population of a city is growing according to the exponential model $p = ce^{kt}$, where $p$ is the population in thousands and $t$ is measured in years. if the population doubles every 11 years, what is $k$, the city’s growth rate? round answer to the nearest hundredth.
a 2.8%
b 4.4%
c 6.3%
d 8.9%

Explanation:

Step1: Understand the exponential model

The exponential growth model is \( P = Ce^{kt} \), where \( C \) is the initial population, \( k \) is the growth rate, \( t \) is time, and \( P \) is the population at time \( t \). We know the population doubles every 11 years, so when \( t = 11 \), \( P = 2C \).

Step2: Substitute values into the model

Substitute \( P = 2C \) and \( t = 11 \) into \( P = Ce^{kt} \):

$$ 2C = Ce^{k \cdot 11} $$

Divide both sides by \( C \) (assuming \( C
eq 0 \)):

$$ 2 = e^{11k} $$

Step3: Solve for \( k \)

Take the natural logarithm of both sides:

$$ \ln(2) = \ln(e^{11k}) $$

Using the property \( \ln(e^x) = x \), we get:

$$ \ln(2) = 11k $$

Then solve for \( k \):

$$ k = \frac{\ln(2)}{11} $$

Calculate \( \ln(2) \approx 0.6931 \), so:

$$ k \approx \frac{0.6931}{11} \approx 0.0630 = 6.30\% $$

Answer:

C. 6.30%