The solution of the logarithmic equation 
𝒍𝒏 𝟐𝒙 − 𝟑 + 𝒍𝒏 𝒙 − 𝟐 = 𝟐𝒍𝒏 𝒙 is:

To solve the logarithmic equation \( \ln(2x - 3) + \ln(x - 2) = 2\ln(x) \), let's proceed step by step.

### Step 1: Use the logarithmic property
We can combine the left-hand side (LHS) using the logarithmic property:
\[
\ln(a) + \ln(b) = \ln(a \cdot b)
\]
Thus, the equation becomes:
\[
\ln\left[(2x - 3)(x - 2)\right] = 2\ln(x)
\]

### Step 2: Simplify the equation
Using the property \( a \ln(b) = \ln(b^a) \) on the right-hand side, we get:
\[
\ln\left[(2x - 3)(x - 2)\right] = \ln(x^2)
\]

### Step 3: Exponentiate both sides
To eliminate the logarithms, exponentiate both sides:
\[
(2x - 3)(x - 2) = x^2
\]

### Step 4: Expand and simplify
Now, expand the left-hand side:
\[
(2x - 3)(x - 2) = 2x^2 - 4x - 3x + 6 = 2x^2 - 7x + 6
\]
Thus, the equation becomes:
\[
2x^2 - 7x + 6 = x^2
\]

### Step 5: Move all terms to one side
Subtract \( x^2 \) from both sides:
\[
2x^2 - 7x + 6 - x^2 = 0
\]
This simplifies to:
\[
x^2 - 7x + 6 = 0
\]

### Step 6: Solve the quadratic equation
Factor the quadratic equation:
\[
x^2 - 7x + 6 = (x - 6)(x - 1) = 0
\]
Thus, the solutions are:
\[
x = 6 \quad \text{or} \quad x = 1
\]

### Step 7: Check for extraneous solutions
Since the original equation involves logarithms, we must check that the solutions satisfy the domain restrictions. The arguments of the logarithms must be positive:
- For \( \ln(2x - 3) \), we require \( 2x - 3 > 0 \), which gives \( x > \frac{3}{2} \).
- For \( \ln(x - 2) \), we require \( x - 2 > 0 \), which gives \( x > 2 \).

Therefore, \( x = 1 \) is **not** a valid solution because it does not satisfy \( x > 2 \). Thus, the only valid solution is:

\[
\boxed{x = 6}
\]

Would you like more details on any step or have further questions?

### Related questions:
1. How do you verify the solution for logarithmic equations involving more variables?
2. What are the domain restrictions for logarithmic functions, and why are they important?
3. How can logarithmic properties simplify solving equations?
4. How do extraneous solutions arise in logarithmic equations?
5. What other types of equations require similar checks for domain restrictions?

**Tip**: Always check the domain restrictions when working with logarithms, as not all solutions from algebraic manipulation will be valid for the original equation.

The solution of the logarithmic equation ln(2x - 3) + ln(x - 2) = 2ln(x) is:

This detailed step-by-step guide explains how to solve the logarithmic equation ln(2x - 3) + ln(x - 2) = 2ln(x). It covers logarithmic properties, algebraic manipulation, and quadratic equation solving. The solution also checks for domain restrictions to avoid extraneous solutions. Suitable for students in Grades 10-12.

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