Given the equation:

\[
\ln(y-1) - \ln(1) = x + \ln(x)
\]

Since \(\ln(1) = 0\), the equation simplifies to:

\[
\ln(y-1) = x + \ln(x)
\]

To solve for \(y\), we exponentiate both sides to eliminate the natural logarithm:

\[
y - 1 = e^{x + \ln(x)}
\]

We can simplify the right-hand side using the property \(e^{a+b} = e^a \cdot e^b\):

\[
y - 1 = e^x \cdot e^{\ln(x)} = e^x \cdot x
\]

Thus, the equation becomes:

\[
y - 1 = x \cdot e^x
\]

Finally, solve for \(y\):

\[
y = x \cdot e^x + 1
\]

So, the solution is:

\[
\boxed{y = x \cdot e^x + 1}
\]

Would you like more details on this solution, or do you have any questions?

Here are 8 related questions you might consider:

1. What are the properties of logarithms used in this problem?
2. How do you differentiate \(y = x \cdot e^x + 1\) with respect to \(x\)?
3. What is the inverse function of \(y = x \cdot e^x + 1\)?
4. How do you solve an equation involving natural logarithms in general?
5. What happens to \(y\) as \(x\) approaches infinity in this equation?
6. Can this equation be solved graphically, and what would it look like?
7. How do you interpret the solution \(y = x \cdot e^x + 1\) in a real-world context?
8. What are the domains and ranges of the functions involved in this equation?

**Tip:** When dealing with logarithmic equations, always remember to check the domain of the original equation to ensure all logarithmic arguments are positive.

Explore how to solve the equation ln(y-1) = x + ln(x) step-by-step. This problem involves logarithms, exponential functions, and equations. Suitable for college-level students and those studying advanced mathematics.

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