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 $k$. $k=\ln (1.026)$ (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. $y(t)=105,000 e^{\ln (1.026) t}$ (Type an expression using $t$ as the variable. Type an exact answer.) b. In what year will the population be 175,000 ? (Round down to the nearest year.)

Step 1 ：Given that the population grows at a rate of 2.6% per year, this is equivalent to a growth factor of 1.026. The rate constant k in an exponential growth function is the natural logarithm of the growth factor. So, we can calculate k as \(\ln(1.026)\).

Step 2 ：Using the calculated value of k, we can devise an exponential growth function. The general form of an exponential growth function is \(y(t) = P_0 \times e^{kt}\), where \(P_0\) is the initial population, k is the rate constant, and t is time. In this case, \(P_0\) is 105,000, k is \(\ln(1.026)\), and t is the number of years since 2016. So the function is \(y(t) = 105,000 \times e^{\ln(1.026)t}\).

Step 3 ：To find the year when the population will reach 175,000, we need to solve the equation \(175,000 = 105,000 \times e^{\ln(1.026)t}\) for t. This will give us the number of years since 2016 when the population will reach 175,000.

Step 4 ：Adding the calculated number of years to 2016 will give us the actual year when the population will reach 175,000.

Step 5 ：Final Answer: The rate constant k is approximately \(\boxed{0.02567}\) and the population will reach 175,000 in the year \(\boxed{2036}\).