The sum of all the roots of the equation `log_(2)(x-1)+log_(2)(x+2)-log_(2)(3x-1)=log_(2)4`

Options

A. 12
B. 2
C. 10
D. 11

Answer

To solve the equation \( \log_{2}(x-1) + \log_{2}(x+2) - \log_{2}(3x-1) = \log_{2}(4) \), we can follow these steps:

### Step 1: Combine the logarithmic expressions
Using the properties of logarithms, we can combine the left-hand side:
\[ \log_{2}(x-1) + \log_{2}(x+2) - \log_{2}(3x-1) = \log_{2}\left(\frac{(x-1)(x+2)}{3x-1}\right) \]
Thus, we rewrite the equation as:
\[ \log_{2}\left(\frac{(x-1)(x+2)}{3x-1}\right) = \log_{2}(4) \]

### Step 2: Set the arguments equal to each other
Since the logarithm function is one-to-one, we can set the arguments equal:
\[ \frac{(x-1)(x+2)}{3x-1} = 4 \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ (x-1)(x+2) = 4(3x-1) \]

### Step 4: Expand both sides
Expanding both sides, we have:
\[ x^2 + 2x - x - 2 = 12x - 4 \]
This simplifies to:
\[ x^2 + x - 2 = 12x - 4 \]

### Step 5: Rearrange the equation
Bringing all terms to one side results in:
\[ x^2 + x - 12x + 4 - 2 = 0 \]
This simplifies to:
\[ x^2 - 11x + 2 = 0 \]

### Step 6: Use the quadratic formula to find the roots
The sum of the roots for a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( -\frac{b}{a} \). Here, \( a = 1 \), \( b = -11 \), and \( c = 2 \):
\[ \text{Sum of roots} = -\frac{-11}{1} = 11 \]

### Final Result
The sum of all the roots of the equation is \( \boxed{11} \).

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