The number of solutions of `"log"_(2) (x-1) = 2 "log"_(2) (x-3)` is

Options

A. 2
B. 1
C. 6
D. 7

Answer

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

### Step 1: Rewrite the equation using properties of logarithms
Using the property of logarithms that states \( a \log_b(c) = \log_b(c^a) \), we can rewrite the right side of the equation:
\[ \log_{2}(x-1) = \log_{2}((x-3)^2) \]

### Step 2: Set the arguments of the logarithms equal to each other
Since the logarithmic function is one-to-one, we can set the arguments equal to each other:
\[ x - 1 = (x - 3)^2 \]

### Step 3: Expand the right side
Now, we will expand the right side of the equation:
\[ x - 1 = x^2 - 6x + 9 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ 0 = x^2 - 6x + 9 - x + 1 \]
\[ 0 = x^2 - 7x + 10 \]

### Step 5: Factor the quadratic equation
Next, we will factor the quadratic equation:
\[ 0 = (x - 5)(x - 2) \]

### Step 6: Solve for \( x \)
Setting each factor to zero gives us the potential solutions:
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

### Step 7: Check for validity of the solutions
We need to check if these solutions are valid in the context of the logarithmic functions:
- For \( x = 2 \):
\[ \log_{2}(2 - 1) = \log_{2}(1) = 0 \]
\[ 2 \log_{2}(2 - 3) = 2 \log_{2}(-1) \quad \text{(undefined)} \]
Thus, \( x = 2 \) is not a valid solution.

- For \( x = 5 \):
\[ \log_{2}(5 - 1) = \log_{2}(4) = 2 \]
\[ 2 \log_{2}(5 - 3) = 2 \log_{2}(2) = 2 \times 1 = 2 \]
Thus, \( x = 5 \) is a valid solution.

### Conclusion
The number of valid solutions to the equation \( \log_{2}(x-1) = 2 \log_{2}(x-3) \) is **1** (only \( x = 5 \)).

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