To solve this problem, we can model the population decline using the formula for exponential decay:

$P(t) = P_0 \times (1 - r)^t$

Where:

• $P(t)$ is the population after $t$ years,
• $P_0$ is the initial population (16,000),
• $r$ is the rate of decrease (2% = 0.02),
• $t$ is the number of years.

We want to find $t$ such that the population falls below 10,000. So, we need to solve for $t$ when $P(t) = 10,000$:

$10,000 = 16,000 \times (1 - 0.02)^t$

Simplifying the equation:

$10,000 = 16,000 \times (0.98)^t$

Dividing both sides by 16,000:

$\frac{10,000}{16,000} = (0.98)^t$

$0.625 = (0.98)^t$

Now, take the natural logarithm (ln) of both sides to solve for $t$:

$\ln(0.625) = t \times \ln(0.98)$

Finally, solve for $t$:

$t = \frac{\ln(0.625)}{\ln(0.98)}$

Let me calculate that.It will take approximately 23.26 years for the town's population to fall below 10,000.

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. How would the time change if the population decrease rate were 3% per year?
2. What if the population started at 20,000? How long would it take to fall below 10,000 at a 2% annual decrease?
3. How would the calculation differ if the rate of decrease were compounded monthly instead of yearly?
4. If the town wanted to maintain a population above 12,000, how many years would it take before reaching this level?
5. What population will the town have after 10 years with the same 2% annual decrease?

Tip: When dealing with exponential growth or decay, logarithms are helpful for solving equations where the variable is an exponent.