The population of a town with a 2016 population of 105,000 grows at a rate of $2.6 \%$ per year. a. Find the rate constant $k$ and use it to devise an exponential growth function that fits the given data. b. In what year will the population reach 175,000 ? a. Find the rate constant $k$. $\mathrm{k}=\ln (1.026)$ (Type an exact answer.) Devise an exponential growth function that fits the given data. Use the value of $\mathrm{k}$ found in the previous step. $y(t)=$ (Type an expression using $t$ as the variable. Type an exact answer.)

Step 1 ：The general form of an exponential growth function is \(y(t) = y_0 * e^{kt}\), where \(y_0\) is the initial amount, \(k\) is the rate constant, and \(t\) is the time. In this case, the initial population is 105,000 and the growth rate is 2.6% per year.

Step 2 ：We can convert the percentage to a decimal and use it to find the rate constant \(k\) by taking the natural logarithm of 1 plus the growth rate. So, \(k = \ln(1.026)\), which is approximately 0.02567.

Step 3 ：Substitute the values into the exponential growth function, we get \(y(t) = 105000 * e^{0.02567t}\).

Step 4 ：To find out in what year will the population reach 175,000, we set \(y(t) = 175000\) and solve for \(t\).

Step 5 ：By solving the equation, we get \(t\) is approximately 19.901459554267042. Since the initial year is 2016, we add \(t\) to 2016 and get the year is approximately 2035.9014595542671.

Step 6 ：Rounding up to the nearest whole year, the population will reach 175,000 in the year 2036.

Step 7 ：So, the final answers are \(k \approx \boxed{0.02567}\), \(y(t) = 105000 * e^{0.02567t}\), and the year is \(\boxed{2036}\).