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

Options

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

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 in exponential form.
Using the property of logarithms, we can rewrite the equation:
\[ x^3 - 3x^2 - 4x + 8 = (x - 3)^3 \]

### Step 2: Expand the right-hand 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 set the left-hand side equal to the expanded right-hand side:
\[ x^3 - 3x^2 - 4x + 8 = x^3 - 9x^2 + 27x - 27 \]
Subtract \( x^3 \) from both sides:
\[ -3x^2 - 4x + 8 = -9x^2 + 27x - 27 \]
Now, move all terms to one side:
\[ -3x^2 + 9x^2 - 4x - 27x + 8 + 27 = 0 \]
This simplifies to:
\[ 6x^2 - 31x + 35 = 0 \]

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

### Step 5: Check the validity of the solutions.
We need to check if these solutions satisfy the conditions for the logarithm:
- The base \( x - 3 \) must be greater than 0, which means \( x > 3 \).
- Therefore, we discard \( x = \frac{5}{3} \) since it is less than 3.

The only valid solution is \( x = \frac{7}{2} \).

### Conclusion
The number of solutions to the equation is **1**.

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