We are given the equation:

$(x+1) \ln 2 = (x-1) \ln 6$

Step 1: Expand both sides

Expand both sides of the equation to make the terms more manageable:

$x \ln 2 + \ln 2 = x \ln 6 - \ln 6$

Step 2: Isolate the terms involving $x$

Move all terms involving $x$ to one side and the constant terms to the other side:

$x \ln 2 - x \ln 6 = -\ln 2 - \ln 6$

Step 3: Factor out $x$

Factor $x$ on the left-hand side:

$x(\ln 2 - \ln 6) = -\ln 2 - \ln 6$

Step 4: Simplify the logarithmic expressions

Use the logarithmic property $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$:

$x \ln \left(\frac{2}{6}\right) = -(\ln 2 + \ln 6)$

Simplify further:

$x \ln \left(\frac{1}{3}\right) = -\ln (2 \cdot 6)$

$x \ln \left(\frac{1}{3}\right) = -\ln 12$

Step 5: Solve for $x$

Divide both sides by $\ln \left(\frac{1}{3}\right)$:

$x = \frac{-\ln 12}{\ln \left(\frac{1}{3}\right)}$

We can simplify $\ln \left(\frac{1}{3}\right)$ as $-\ln 3$, so:

$x = \frac{-\ln 12}{-\ln 3} = \frac{\ln 12}{\ln 3}$

Finally, use the logarithmic property $\frac{\ln a}{\ln b} = \log_b a$:

$x = \log_3 12$

Thus, the solution is:

$x = \log_3 12$

Conclusion:

The solution to the equation is $x = \log_3 12$.

Would you like more details or have any questions about the solution?

Here are 5 related questions to explore:

1. How would you solve $\log_2 (x+1) = \log_3 (x-1)$?
2. What is the value of $x$ if you approximate $\log_3 12$?
3. How does changing the base of the logarithm affect the solution process?
4. What are the properties of logarithms that simplify solving such equations?
5. Can you solve $\ln(2x + 1) = \ln(6x - 2)$ for $x$?

Tip: Always remember the properties of logarithms like $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$, which are crucial for simplifying and solving logarithmic equations.