\[ P=312(1.04)^{\frac{1}{4}} \] The equation above can be used to model the population, in thousands, of a certain city \( t \) years after 1980 . According to the model, the population is predicted to increase by \( 4 \% \) every \( n \) months. What is the value of \( n \) ?

To determine the value of \( n \) such that the population increases by \( 4\% \) every \( n \) months, we start with the given equation: \[ P = 312(1.04)^{\frac{1}{4}} \] This equation indicates that the population increases by a factor of \( 1.04 \) every \( 4 \) years, since the exponent \( \frac{1}{4} \) corresponds to \( t \) being in years. To find the equivalent monthly increase, we need to express the annual growth rate in terms of months. Since there are \( 12 \) months in a year, we can convert the annual growth factor to a monthly growth factor. Let \( r \) be the monthly growth factor. The relationship between the annual growth factor and the monthly growth factor can be expressed as: \[ (1 + r)^{12} = 1.04 \] To find \( r \), we take the twelfth root of \( 1.04 \): \[ 1 + r = (1.04)^{\frac{1}{12}} \] Now, we can calculate \( r \): \[ r = (1.04)^{\frac{1}{12}} - 1 \] Next, we need to find \( n \) such that the population increases by \( 4\% \) every \( n \) months. This means we want: \[ (1 + r)^n = 1.04 \] Substituting \( 1 + r \): \[ ((1.04)^{\frac{1}{12}})^n = 1.04 \] This simplifies to: \[ (1.04)^{\frac{n}{12}} = 1.04 \] Taking the logarithm of both sides or recognizing that the bases are the same, we can equate the exponents: \[ \frac{n}{12} = 1 \] Solving for \( n \): \[ n = 12 \] Thus, the value of \( n \) is: \[ \boxed{12} \] This means the population is predicted to increase by \( 4\% \) every \( 12 \) months (or every year).