If log x + log y = log (x + y) then :

Options

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

Answer

To solve the equation \( \log x + \log y = \log (x + y) \), we will use properties of logarithms.

### Step 1: Use the property of logarithms
We know that \( \log a + \log b = \log(ab) \). Therefore, we can rewrite the left-hand side of the equation:

\[ \log x + \log y = \log(xy) \]

### Step 2: Set the two logarithmic expressions equal
Now we can set the two expressions equal to each other:

\[ \log(xy) = \log(x + y) \]

### Step 3: Exponentiate both sides
To eliminate the logarithm, we can exponentiate both sides. This gives us:

\[ xy = x + y \]

### Step 4: Rearrange the equation
Now we can rearrange the equation to bring all terms to one side:

\[ xy - x - y = 0 \]

### Step 5: Factor the equation
Next, we can factor the left-hand side:

\[ xy - x - y + 1 = 1 \]

This can be factored as:

\[ (x - 1)(y - 1) = 1 \]

### Step 6: Analyze the factored form
From the factored form \( (x - 1)(y - 1) = 1 \), we can see that \( x - 1 \) and \( y - 1 \) are two numbers whose product is 1. This gives us several possibilities for \( x \) and \( y \).

### Conclusion
The solutions can be expressed in terms of \( x \) and \( y \) as follows:

1. \( x = y \)
2. \( xy = 1 \)
3. \( y = x - 1 \)
4. \( y = \frac{x}{x - 1} \)