 1. If 𝑝(𝑥) = 2^x + 1 and 𝑞(𝑥) = 𝑥^2 − 3 then determine the value(s) of 𝑥 such that 𝑝^-1[𝑞(𝑥)] = 5.

Given:
- \( p(x) = 2^x + 1 \)
- \( q(x) = x^2 - 3 \)

We need to determine the value(s) of \( x \) such that \( p^{-1}(q(x)) = 5 \).

### Solution:

1. **Find \( p^{-1}(x) \):**

   Start with the function \( p(x) = 2^x + 1 \).

   To find the inverse, set \( y = 2^x + 1 \) and solve for \( x \):
   \[
   y - 1 = 2^x
   \]
   Taking the logarithm base 2 on both sides:
   \[
   x = \log_2(y - 1)
   \]
   Therefore, \( p^{-1}(x) = \log_2(x - 1) \).

2. **Substitute \( q(x) \) into \( p^{-1} \):**

   We need to solve the equation \( p^{-1}(q(x)) = 5 \). Substitute \( q(x) = x^2 - 3 \) into \( p^{-1} \):
   \[
   p^{-1}(x^2 - 3) = 5
   \]
   \[
   \log_2(x^2 - 3 - 1) = 5
   \]
   Simplifying inside the logarithm:
   \[
   \log_2(x^2 - 4) = 5
   \]

3. **Solve for \( x \):**

   To eliminate the logarithm, rewrite the equation:
   \[
   x^2 - 4 = 2^5
   \]
   \[
   x^2 - 4 = 32
   \]
   \[
   x^2 = 36
   \]
   \[
   x = \pm 6
   \]

### Conclusion:
The values of \( x \) that satisfy \( p^{-1}(q(x)) = 5 \) are \( x = 6 \) and \( x = -6 \).

Would you like any more details, or do you have any further questions?

### Follow-up Questions:
1. How does the logarithmic property \( \log_b(x) = y \) translate graphically?
2. What are the conditions for \( p(x) \) to have an inverse function?
3. How would the solution change if \( q(x) \) were a cubic function?
4. Can \( p(x) = 2^x + 1 \) ever be a decreasing function?
5. What is the geometric interpretation of finding the inverse function?

### Tip:
When dealing with logarithmic equations, ensure that the argument inside the logarithm is positive since logarithms of non-positive numbers are undefined in the real number system.

Explore how to solve the inverse function problem involving exponential and logarithmic functions. Learn the steps to find the values of x that satisfy p^-1(q(x)) = 5. This problem covers key concepts in exponential and logarithmic functions, suitable for advanced high school students.

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