Final Answer:
a) The time when the population of Mooresville is 192, \( t = \frac{{\log(192)}}{{\log(12.2)}} \), approximately \( t \approx 4.05 \) years.
b) After 4 years, the population of Mooresville will be \( p = 12.2^4 \), approximately \( p \approx 238.52 \) thousand.
Step-by-step explanation:
In part (a), to find the time (\( t \)) when the population is 192, we use the given population model \( p = 12.2^t \). We set \( p \) equal to 192 and solve for \( t \):
\[ 192 = 12.2^t \]
Taking the logarithm base 12.2 on both sides:
\[ \log_{12.2}(192) = t \]
Thus, \( t \) is approximately \( \frac{{\log(192)}}{{\log(12.2)}} \), which evaluates to \( t \approx 4.05 \) years.
In part (b), to find the population (\( p \)) after 4 years, we substitute \( t = 4 \) into the population model:
\[ p = 12.2^4 \]
Evaluating this expression gives \( p \approx 238.52 \) thousand.
Therefore, after 4 years, the population of Mooresville will be approximately 238.52 thousand.
These calculations are based on the exponential growth model provided, where the population (\( p \)) is modeled as \( 12.2^t \), with \( t \) representing the number of years. The logarithmic calculation in part (a) allows us to find the specific time when the population reaches 192, while part (b) involves straightforward substitution to determine the population after 4 years.