Given the function $f(x) = 3 + x + \ln(x - 2)$, where $x > 2$, let's go through the two parts of the question.

Part (a) Find $f^{-1}(6)$

To find $f^{-1}(6)$, we want to determine the value of $x$ such that $f(x) = 6$.

1. Start with the equation: $3 + x + \ln(x - 2) = 6$
2. Subtract 3 from both sides: $x + \ln(x - 2) = 3$
3. We need to solve this equation for $x$. This is a transcendental equation, meaning it involves both algebraic and logarithmic terms, and might require numerical methods.

Part (b) Find $f^{-1}(f(16))$

We know that $f^{-1}(f(16)) = 16$, because the inverse function applied to the function itself returns the original input value.

Thus: $f^{-1}(f(16)) = 16$

Now, let's proceed by solving $x + \ln(x - 2) = 3$ to get $f^{-1}(6)$. I'll calculate this next.The solution to $f^{-1}(6)$ is $x = 3$.

Final answers:

(a) $f^{-1}(6) = 3$

(b) $f^{-1}(f(16)) = 16$

Would you like further details on the process, or do you have any questions?

Here are five follow-up questions:

1. How do we find the inverse of a logarithmic function in general?
2. Why does $f^{-1}(f(x)) = x$ hold for any value of $x$?
3. What are transcendental equations, and why are they hard to solve algebraically?
4. What is the domain of $f^{-1}(x)$ in this problem?
5. How do numerical methods like $fsolve$ work?

Tip: When solving transcendental equations involving logarithms, a good initial estimate helps numerical solvers converge faster.