if log (x+y) = log x + log y , then x= ____

Options

A. `(-y)/(1=y)`
B. `(y)/(y-1)`
C. 1
D. `y/(1+y)`

Explanation

To solve the equation \( \log(x+y) = \log x + \log y \), we will follow these steps:

### Step 1: Use the property of logarithms
We know that \( \log a + \log b = \log(ab) \). Therefore, we can rewrite the right side of the equation:
\[ \log(x+y) = \log(xy) \]

### Step 2: Remove the logarithm
Since the logarithm function is one-to-one, we can set the arguments equal to each other:
\[ x + y = xy \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ xy - x - y = 0 \]

### Step 4: Factor the equation
We can factor the left-hand side:
\[ x(y - 1) - y = 0 \]
This can be rearranged to:
\[ x(y - 1) = y \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \) by dividing both sides by \( (y - 1) \) (assuming \( y \neq 1 \)):
\[ x = \frac{y}{y - 1} \]

### Conclusion
Thus, the value of \( x \) is:
\[ x = \frac{y}{y - 1} \]

---