If `log(x-y)-log5-1/2logx-1/2logy=0` then `x/y+y/x` is equal to

Options

A. 25
B. 26
C. 27
D. 28

Answer

To solve the equation \( \log(x - y) - \log 5 - \frac{1}{2} \log x - \frac{1}{2} \log y = 0 \) and find the value of \( \frac{x}{y} + \frac{y}{x} \), we will follow these steps:

### Step 1: Rewrite the equation using logarithmic properties
We start with the original equation:
\[ \log(x - y) - \log 5 - \frac{1}{2} \log x - \frac{1}{2} \log y = 0 \]
We can use the property of logarithms that states \( \frac{1}{2} \log a = \log a^{1/2} \) to rewrite the equation:
\[ \log(x - y) - \log 5 - \log(x^{1/2}) - \log(y^{1/2}) = 0 \]

### Step 2: Combine the logarithmic terms
Using the property \( \log a - \log b = \log \frac{a}{b} \), we can combine the logarithmic terms:
\[ \log(x - y) - \log(5 \sqrt{xy}) = 0 \]
This simplifies to:
\[ \log \frac{x - y}{5 \sqrt{xy}} = 0 \]

### Step 3: Exponentiate both sides
Exponentiating both sides gives us:
\[ \frac{x - y}{5 \sqrt{xy}} = 1 \]
This leads to:
\[ x - y = 5 \sqrt{xy} \]

### Step 4: Rearrange the equation
Rearranging the equation, we have:
\[ x - y - 5 \sqrt{xy} = 0 \]

### Step 5: Square both sides
To eliminate the square root, we square both sides:
\[ (x - y)^2 = (5 \sqrt{xy})^2 \]
This simplifies to:
\[ x^2 - 2xy + y^2 = 25xy \]

### Step 6: Rearrange to form a quadratic equation
Rearranging gives:
\[ x^2 - 27xy + y^2 = 0 \]

### Step 7: Use the quadratic formula
We can treat this as a quadratic in \( x \):
\[ x^2 - 27xy + y^2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -27y, c = y^2 \):
\[ x = \frac{27y \pm \sqrt{(-27y)^2 - 4 \cdot 1 \cdot y^2}}{2 \cdot 1} \]
\[ x = \frac{27y \pm \sqrt{729y^2 - 4y^2}}{2} \]
\[ x = \frac{27y \pm \sqrt{725y^2}}{2} \]
\[ x = \frac{27y \pm 27y\sqrt{\frac{725}{729}}}{2} \]
\[ x = \frac{27y(1 \pm \sqrt{\frac{725}{729}})}{2} \]

### Step 8: Find \( \frac{x}{y} + \frac{y}{x} \)
Using the identity:
\[ \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} \]
From the quadratic equation \( x^2 - 27xy + y^2 = 0 \), we have:
\[ x^2 + y^2 = 27xy \]
Thus:
\[ \frac{x^2 + y^2}{xy} = \frac{27xy}{xy} = 27 \]

### Final Result
Therefore, the value of \( \frac{x}{y} + \frac{y}{x} \) is:
\[ \boxed{27} \]