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.

Solve the logarithmic equation (x+1)ln2 = (x-1)ln6 with detailed steps. Learn how to apply logarithmic properties and the change of base theorem to simplify and solve equations like this. Suitable for high school students in Grades 10-12.

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