Solve : `log_(1-2x)(6x^(2)-5x+1)-log_(1-3x)(4x^(2)-4x+1)=2`

Answer

To solve the equation

\[ \log_{(1-2x)}(6x^2 - 5x + 1) - \log_{(1-3x)}(4x^2 - 4x + 1) = 2, \]

we will follow these steps:

### Step 1: Factor the Quadratics

First, we need to factor the quadratic expressions in the logarithms.

1. For \(6x^2 - 5x + 1\):
\[ 6x^2 - 5x + 1 = (2x - 1)(3x - 1). \]

2. For \(4x^2 - 4x + 1\):
\[ 4x^2 - 4x + 1 = (2x - 1)^2. \]

### Step 2: Rewrite the Equation

Now substituting the factored forms back into the equation, we have:

\[ \log_{(1-2x)}((2x-1)(3x-1)) - \log_{(1-3x)}((2x-1)^2) = 2. \]

### Step 3: Use Logarithmic Properties

Using the properties of logarithms, we can rewrite the equation:

\[ \log_{(1-2x)}((2x-1)(3x-1)) - 2\log_{(1-3x)}(2x-1) = 2. \]

This can be rewritten as:

\[ \log_{(1-2x)}((2x-1)(3x-1)) - \log_{(1-3x)}((2x-1)^2) = 2. \]

### Step 4: Combine the Logarithms

Using the logarithmic identity \(\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)\):

\[ \log_{(1-2x)}\left(\frac{(2x-1)(3x-1)}{(2x-1)^2}\right) = 2. \]

This simplifies to:

\[ \log_{(1-2x)}\left(\frac{3x-1}{2x-1}\right) = 2. \]

### Step 5: Convert Logarithmic Equation to Exponential Form

From the logarithmic equation, we can write:

\[ \frac{3x-1}{2x-1} = (1-2x)^2. \]

### Step 6: Solve the Resulting Equation

Cross-multiplying gives:

\[ 3x - 1 = (1 - 2x)^2(2x - 1). \]

Expanding the right side:

\[ (1 - 2x)^2 = 1 - 4x + 4x^2, \]
\[ (1 - 2x)^2(2x - 1) = (1 - 4x + 4x^2)(2x - 1). \]

Expanding this product leads to:

\[ = 2x - 8x^2 + 8x^3 - 1 + 4x - 4x^2, \]
\[ = 8x^3 - 12x^2 + 6x - 1. \]

Setting the equation \(3x - 1 = 8x^3 - 12x^2 + 6x - 1\) leads to:

\[ 0 = 8x^3 - 12x^2 + 3x. \]

Factoring out \(x\):

\[ x(8x^2 - 12x + 3) = 0. \]

### Step 7: Solve for \(x\)

This gives us \(x = 0\) or solving the quadratic \(8x^2 - 12x + 3 = 0\) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 8 \cdot 3}}{2 \cdot 8} = \frac{12 \pm \sqrt{144 - 96}}{16} = \frac{12 \pm \sqrt{48}}{16} = \frac{12 \pm 4\sqrt{3}}{16} = \frac{3 \pm \sqrt{3}}{4}. \]

### Step 8: Check Validity of Solutions

We need to check which of these solutions satisfy the conditions for the logarithms:

1. \(1 - 2x > 0\) implies \(x < \frac{1}{2}\).
2. \(1 - 3x > 0\) implies \(x < \frac{1}{3}\).
3. \(2x - 1 \neq 0\) implies \(x \neq \frac{1}{2}\).

Thus, the valid solution from \(x = 0\), \(x = \frac{3 + \sqrt{3}}{4}\), and \(x = \frac{3 - \sqrt{3}}{4}\) is \(x = \frac{3 - \sqrt{3}}{4}\) since it is the only one that satisfies all conditions.

### Final Answer

Thus, the solution to the equation is:

\[ \boxed{\frac{3 - \sqrt{3}}{4}}. \]