The population P of a small town in hundreds, is  modeled by the inverse of the function P^-1(t)=20ln(40t-1200), where t is measured in years. Find the equation to model the population 

To find the equation that models the population \( P \) based on the given inverse function, we can proceed as follows:

Given:

\[
P^{-1}(t) = 20 \ln(40t - 1200)
\]

This function \( P^{-1}(t) \) is the inverse of the function that models the population \( P \). To find \( P \) as a function of \( t \), we need to express \( t \) in terms of \( P \) from the given equation and then solve for \( P \).

1. **Express \( t \) in terms of \( P \):**

   We have:

   \[
   t = P^{-1}(t) = 20 \ln(40t - 1200)
   \]

   To find the function \( P(t) \), solve the equation for \( t \) in terms of \( P \):

   \[
   \frac{t}{20} = \ln(40t - 1200)
   \]

2. **Solve for \( t \):**

   Let's use the inverse operation to get rid of the natural logarithm. Let:

   \[
   P = 20 \ln(40t - 1200)
   \]

   Divide both sides by 20:

   \[
   \frac{P}{20} = \ln(40t - 1200)
   \]

   Exponentiate both sides to solve for \( t \):

   \[
   e^{\frac{P}{20}} = 40t - 1200
   \]

   Add 1200 to both sides:

   \[
   40t = 1200 + e^{\frac{P}{20}}
   \]

   Finally, solve for \( t \):

   \[
   t = \frac{1200 + e^{\frac{P}{20}}}{40}
   \]

3. **Find the inverse:**

   Since we have \( P^{-1}(t) \), we want the function \( P(t) \) that gives us the population based on the year \( t \). Rearrange the equation above to solve for \( P \) in terms of \( t \):

   From \( t = \frac{1200 + e^{\frac{P}{20}}}{40} \), solve for \( P \):

   \[
   40t = 1200 + e^{\frac{P}{20}}
   \]

   Subtract 1200 from both sides:

   \[
   40t - 1200 = e^{\frac{P}{20}}
   \]

   Take the natural logarithm of both sides:

   \[
   \ln(40t - 1200) = \frac{P}{20}
   \]

   Multiply both sides by 20 to solve for \( P \):

   \[
   P = 20 \ln(40t - 1200)
   \]

Therefore, the equation that models the population \( P \) in hundreds as a function of time \( t \) in years is:

\[
\boxed{P(t) = 20 \ln(40t - 1200)}
\]

Do you have any more questions, or would you like further details on any part of the solution?

### Related Questions

1. How do you find the domain of the function \( P(t) = 20 \ln(40t - 1200) \)?
2. What is the range of the function \( P(t) = 20 \ln(40t - 1200) \)?
3. How does the population change over time according to the model \( P(t) \)?
4. How would the model change if the function were \( P^{-1}(t) = 25 \ln(30t - 900) \)?
5. What are the implications of the population model in terms of real-world population growth?

### Tip

When working with inverse functions, remember that solving for one variable in terms of another can often require using logarithmic and exponential properties to reverse operations correctly.

Learn how to model population growth of a small town using inverse functions and natural logarithms. Explore the equation P(t) = 20 ln(40t - 1200) derived from the given inverse function P^-1(t). Suitable for high school students studying algebra and functions.

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