The number of solutions of `log_(4)(x-1)=log_(2)(x-3)` is

Options

A. `3`
B. `1`
C. `2`
D. `0`

Answer

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

### Step 1: Rewrite the logarithms
We can express \( \log_{4}(x-1) \) in terms of base 2:
\[ \log_{4}(x-1) = \frac{\log_{2}(x-1)}{\log_{2}(4)} = \frac{\log_{2}(x-1)}{2} \]
So, we can rewrite the equation as:
\[ \frac{\log_{2}(x-1)}{2} = \log_{2}(x-3) \]

### Step 2: Eliminate the fraction
Multiply both sides by 2 to eliminate the fraction:
\[ \log_{2}(x-1) = 2 \log_{2}(x-3) \]

### Step 3: Use the power property of logarithms
Using the property \( n \log_{b}(a) = \log_{b}(a^n) \), we can rewrite the right side:
\[ \log_{2}(x-1) = \log_{2}((x-3)^2) \]

### Step 4: Set the arguments equal
Since the logarithms are equal, we can set the arguments equal to each other:
\[ x-1 = (x-3)^2 \]

### Step 5: Expand the equation
Expanding the right side gives:
\[ x - 1 = x^2 - 6x + 9 \]

### Step 6: Rearrange the equation
Rearranging the equation to one side:
\[ 0 = x^2 - 6x - x + 9 + 1 \]
\[ 0 = x^2 - 7x + 10 \]

### Step 7: Factor the quadratic
Factoring the quadratic equation:
\[ 0 = (x - 5)(x - 2) \]

### Step 8: 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 9: Check the validity of solutions
We need to check if these solutions satisfy the conditions for logarithms:
1. For \( x = 5 \):
- \( x - 1 = 4 > 0 \)
- \( x - 3 = 2 > 0 \) (valid)

2. For \( x = 2 \):
- \( x - 1 = 1 > 0 \)
- \( x - 3 = -1 < 0 \) (invalid)

### Conclusion
The only valid solution is \( x = 5 \). Therefore, the number of solutions to the equation \( \log_{4}(x-1) = \log_{2}(x-3) \) is **1**.

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