Let x, y be real numbers such that `x gt 2y gt 0` and
`2log (x-2y) = log x + log y`.
Then the possible values (s) of `(x)/(y)`

Options

A. is 1 only
B. are 1 and 4
C. is 4 only
D. is 8 only

Answer

To solve the problem, we need to find the possible values of \( \frac{x}{y} \) given the conditions \( x > 2y > 0 \) and the equation \( 2 \log(x - 2y) = \log x + \log y \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ 2 \log(x - 2y) = \log x + \log y \]

2. **Use the properties of logarithms:**
The right-hand side can be combined using the product property of logarithms:
\[ \log x + \log y = \log(xy) \]
Therefore, we can rewrite the equation as:
\[ 2 \log(x - 2y) = \log(xy) \]

3. **Exponentiate both sides:**
To eliminate the logarithm, we exponentiate both sides:
\[ (x - 2y)^2 = xy \]

4. **Expand the left-hand side:**
\[ x^2 - 4xy + 4y^2 = xy \]

5. **Rearrange the equation:**
Move all terms to one side:
\[ x^2 - 5xy + 4y^2 = 0 \]

6. **This is a quadratic equation in \( x \):**
We can apply the quadratic formula where \( a = 1, b = -5y, c = 4y^2 \):
\[ x = \frac{-(-5y) \pm \sqrt{(-5y)^2 - 4 \cdot 1 \cdot 4y^2}}{2 \cdot 1} \]
Simplifying this gives:
\[ x = \frac{5y \pm \sqrt{25y^2 - 16y^2}}{2} \]
\[ x = \frac{5y \pm \sqrt{9y^2}}{2} \]
\[ x = \frac{5y \pm 3y}{2} \]

7. **Calculate the two possible values for \( x \):**
- First case:
\[ x = \frac{8y}{2} = 4y \]
- Second case:
\[ x = \frac{2y}{2} = y \]

8. **Evaluate the conditions:**
We have two cases for \( x \):
- If \( x = 4y \), then \( \frac{x}{y} = 4 \).
- If \( x = y \), then \( \frac{x}{y} = 1 \).

9. **Check the inequality \( x > 2y \):**
- For \( x = 4y \): \( 4y > 2y \) (valid)
- For \( x = y \): \( y > 2y \) (invalid)

Thus, the only valid solution is:
\[ \frac{x}{y} = 4 \]

### Final Answer:
The possible value of \( \frac{x}{y} \) is \( 4 \).