The given equation in the image appears to be:

$\ln(x+1) + \ln(x-1) - 3 = 0$

We are tasked with solving this equation.

Steps to Solve:

1. Combine the logarithmic terms using the product rule: Recall that $\ln(a) + \ln(b) = \ln(a \cdot b)$. Applying this: $\ln(x+1) + \ln(x-1) = \ln((x+1)(x-1))$ So the equation becomes: $\ln((x+1)(x-1)) - 3 = 0$

2. Simplify: The expression inside the logarithm simplifies as: $(x+1)(x-1) = x^2 - 1$ Therefore, the equation becomes: $\ln(x^2 - 1) - 3 = 0$

3. Isolate the logarithmic term: Add 3 to both sides: $\ln(x^2 - 1) = 3$

4. Rewrite in exponential form: Recall that $\ln(a) = b \implies a = e^b$. Applying this: $x^2 - 1 = e^3$

5. Solve for $x^2$: Add 1 to both sides: $x^2 = e^3 + 1$

6. Solve for $x$: Take the square root of both sides: $x = \pm\sqrt{e^3 + 1}$

   However, note that the domain of the logarithmic terms requires $x+1 > 0$ and $x-1 > 0$, meaning $x > 1$. Therefore, we discard the negative root.

   The final solution is: $x = \sqrt{e^3 + 1}$

Final Answer:

$x = \sqrt{e^3 + 1}$

Would you like a detailed explanation of any step or related questions? Here are five questions that build on this problem:

1. How does the domain of logarithmic functions affect the solution?
2. What is the significance of rewriting logarithmic equations in exponential form?
3. Can you explain the product rule for logarithms in more depth?
4. How would the solution change if the equation involved $\ln(x^2 - 1) + 3 = 0$?
5. What is the numerical approximation of $x$ using $e \approx 2.718$?

Tip: When solving logarithmic equations, always check the domain to ensure the solution is valid!