The number of solutions of the equation `log_(x-3)(x^(3)-3x^(2)-4x+8)=3` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( \log_{(x-3)}(x^3 - 3x^2 - 4x + 8) = 3 \), we will follow these steps:

### Step 1: Rewrite the logarithmic equation
Using the property of logarithms, we can rewrite the equation:
\[ x^3 - 3x^2 - 4x + 8 = (x-3)^3 \]

### Step 2: Expand the right side
Now, we will expand \( (x-3)^3 \):
\[ (x-3)^3 = x^3 - 9x^2 + 27x - 27 \]

### Step 3: Set the equation to zero
Now, we equate both sides:
\[ x^3 - 3x^2 - 4x + 8 = x^3 - 9x^2 + 27x - 27 \]
Subtract \( x^3 \) from both sides:
\[ -3x^2 - 4x + 8 = -9x^2 + 27x - 27 \]
Rearranging gives:
\[ 6x^2 - 31x + 35 = 0 \]

### Step 4: Solve the quadratic equation
Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 6 \), \( b = -31 \), and \( c = 35 \):
\[ x = \frac{31 \pm \sqrt{(-31)^2 - 4 \cdot 6 \cdot 35}}{2 \cdot 6} \]
Calculating the discriminant:
\[ (-31)^2 = 961 \quad \text{and} \quad 4 \cdot 6 \cdot 35 = 840 \]
Thus:
\[ \sqrt{961 - 840} = \sqrt{121} = 11 \]
Now substituting back:
\[ x = \frac{31 \pm 11}{12} \]
Calculating the two possible values:
1. \( x = \frac{42}{12} = \frac{7}{2} = 3.5 \)
2. \( x = \frac{20}{12} = \frac{5}{3} \approx 1.67 \)

### Step 5: Check the validity of solutions
We need to check if these solutions satisfy the condition \( x - 3 > 0 \) (i.e., \( x > 3 \)):
- For \( x = \frac{7}{2} = 3.5 \), it satisfies \( x > 3 \).
- For \( x = \frac{5}{3} \approx 1.67 \), it does not satisfy \( x > 3 \).

### Conclusion
Thus, the only valid solution is \( x = \frac{7}{2} \). Therefore, the number of solutions to the equation is:
\[ \boxed{1} \]

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