If `2log_4 x =1+log_4 (x-1)` , find the value of `x`

Options

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

Answer

To solve the equation \( 2\log_4 x = 1 + \log_4 (x - 1) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2\log_4 x = 1 + \log_4 (x - 1) \]

### Step 2: Convert the constant to logarithmic form
We know that \( 1 \) can be expressed as \( \log_4 4 \). Therefore, we can rewrite the equation:
\[ 2\log_4 x = \log_4 4 + \log_4 (x - 1) \]

### Step 3: Use logarithmic properties
Using the property that \( \log_a b + \log_a c = \log_a (b \cdot c) \), we can combine the logarithms on the right side:
\[ 2\log_4 x = \log_4 (4(x - 1)) \]

### Step 4: Apply the power rule of logarithms
Using the property \( m\log_a b = \log_a (b^m) \), we can rewrite the left side:
\[ \log_4 (x^2) = \log_4 (4(x - 1)) \]

### Step 5: Set the arguments equal
Since the logarithms are equal, we can set the arguments equal to each other:
\[ x^2 = 4(x - 1) \]

### Step 6: Expand and rearrange the equation
Expanding the right side gives:
\[ x^2 = 4x - 4 \]
Rearranging this equation leads to:
\[ x^2 - 4x + 4 = 0 \]

### Step 7: Factor the quadratic equation
The equation \( x^2 - 4x + 4 = 0 \) can be factored as:
\[ (x - 2)^2 = 0 \]

### Step 8: Solve for \( x \)
Setting the factored form equal to zero gives:
\[ x - 2 = 0 \implies x = 2 \]

### Conclusion
The value of \( x \) is:
\[ \boxed{2} \]