To solve this, let's use the exponential decay formula:

$y = y_0 e^{-kt}$

where:

• $y_0 = 82,700$ thousand (the initial population),
• $y = 82,000$ thousand (the target population),
• $k = 0.00032$ (the decay rate),
• $t$ is the time in years we are trying to find.

Let's start by rearranging the formula to solve for $t$:

1. Substitute the values into the formula:

   $82,000 = 82,700 e^{-0.00032t}$

2. Divide both sides by 82,700 to isolate the exponential term:

   $\frac{82,000}{82,700} = e^{-0.00032t}$

3. Calculate the left side:

   $\frac{82,000}{82,700} \approx 0.9915$

4. Now, take the natural logarithm of both sides to solve for $t$:

   $\ln(0.9915) = -0.00032t$

5. Calculate $\ln(0.9915)$:

   $\ln(0.9915) \approx -0.0085$

6. Now, divide by $-0.00032$ to find $t$:

   $t = \frac{-0.0085}{-0.00032} \approx 26.56$

So, it will take approximately 26.56 years for the population to reach 82,000 thousand.

Would you like a more detailed explanation, or do you have any questions?

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Related Questions

1. How would the time change if the decay rate was 0.05% instead of 0.032%?
2. What is the population after 10 years with the same decay rate?
3. If the population started at 100,000 thousand, how long would it take to reach 82,000 thousand?
4. How would the equation change if the population was increasing instead of decreasing?
5. What is the half-life of this population decline at a rate of 0.032% per year?

Tip

Remember, when solving for $t$ in exponential decay, taking the natural logarithm is essential to isolate the variable in the exponent.