If `log_(2)(x-16)=log_(4)(x-4)`, which of the following could be the value of x ?

Options

A. 12
B. 13
C. 16
D. 20

Explanation

To solve the equation \( \log_{2}(x-16) = \log_{4}(x-4) \), we will follow these steps:

### Step 1: Change the base of the logarithm
We can rewrite the logarithm on the right side using the change of base formula:
\[ \log_{4}(x-4) = \frac{\log_{2}(x-4)}{\log_{2}(4)} \]
Since \( \log_{2}(4) = 2 \), we have:
\[ \log_{4}(x-4) = \frac{\log_{2}(x-4)}{2} \]

### Step 2: Set the equation
Now, substituting this back into the original equation gives us:
\[ \log_{2}(x-16) = \frac{1}{2} \log_{2}(x-4) \]

### Step 3: Multiply both sides by 2
To eliminate the fraction, multiply both sides by 2:
\[ 2 \log_{2}(x-16) = \log_{2}(x-4) \]

### Step 4: Use properties of logarithms
Using the property of logarithms, we can rewrite the left side:
\[ \log_{2}((x-16)^2) = \log_{2}(x-4) \]

### Step 5: Set the arguments equal
Since the logarithms are equal, we can set the arguments equal to each other:
\[ (x-16)^2 = x-4 \]

### Step 6: Expand and rearrange the equation
Expanding the left side:
\[ x^2 - 32x + 256 = x - 4 \]
Rearranging gives:
\[ x^2 - 33x + 260 = 0 \]

### Step 7: Factor the quadratic equation
To factor the quadratic equation, we look for two numbers that multiply to \( 260 \) and add to \( -33 \). The factors are \( -20 \) and \( -13 \):
\[ (x - 20)(x - 13) = 0 \]

### Step 8: Solve for x
Setting each factor to zero gives us:
\[ x - 20 = 0 \quad \text{or} \quad x - 13 = 0 \]
Thus, we have:
\[ x = 20 \quad \text{or} \quad x = 13 \]

### Step 9: Check the validity of the solutions
We need to check if these values make the arguments of the logarithms valid:
- For \( x = 20 \):
\[ \log_{2}(20 - 16) = \log_{2}(4) \quad \text{(valid)} \]
\[ \log_{4}(20 - 4) = \log_{4}(16) \quad \text{(valid)} \]

- For \( x = 13 \):
\[ \log_{2}(13 - 16) = \log_{2}(-3) \quad \text{(not valid)} \]

Thus, the only valid solution is \( x = 20 \).

### Final Answer
The value of \( x \) could be \( 20 \).

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