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The image you provided contains the following mathematical problem:

**Given an inverse \( g(x) = \ln(e^x - 2) \), find \( f(x) \) such that \( f(g(x)) = x \) and \( g(f(x)) = x \).**

Let's solve this step by step.

### Step 1: Express \( g(x) \) explicitly.
The function \( g(x) \) is given as:
\[
g(x) = \ln(e^x - 2)
\]

### Step 2: Determine the inverse function \( f(x) \).
We know that \( f(x) \) should be the inverse of \( g(x) \), which means:
\[
f(g(x)) = x \quad \text{and} \quad g(f(x)) = x
\]

Since \( g(x) \) is given, we solve \( y = g(x) \) for \( x \) to find \( f(x) \).

1. Start with \( y = \ln(e^x - 2) \).
2. Exponentiate both sides to remove the natural logarithm:
   \[
   e^y = e^x - 2
   \]
3. Solve for \( x \):
   \[
   e^x = e^y + 2
   \]
   \[
   x = \ln(e^y + 2)
   \]
Thus, the inverse function \( f(x) \) is:
\[
f(x) = \ln(e^x + 2)
\]

### Final Answer:
The function \( f(x) \) that satisfies \( f(g(x)) = x \) and \( g(f(x)) = x \) is:
\[
f(x) = \ln(e^x + 2)
\]

Would you like more details or have any questions?

Here are five related questions to consider:

1. What is the domain and range of the function \( g(x) = \ln(e^x - 2) \)?
2. How do you verify that \( f(x) \) is indeed the inverse of \( g(x) \)?
3. What are the properties of logarithmic functions that were used in finding the inverse?
4. How would the inverse change if the function were \( g(x) = \ln(e^x - a) \) instead of \( g(x) = \ln(e^x - 2) \)?
5. Can you generalize the process of finding inverses for other functions involving logarithms and exponentials?

**Tip:** When finding an inverse function, ensure that the operations you apply can be reversed step by step to return to the original variable.

Given an inverse g(x) = ln(e^x - 2), find f(x) such that f(g(x)) = x and g(f(x)) = x.

Learn how to find the inverse function f(x) for g(x) = ln(e^x - 2) in this detailed step-by-step solution. This problem explores key concepts such as inverse functions, logarithms, and exponential functions, making it suitable for undergraduate students.

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