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.