If `1/2 log x + 1/2 log y + log2 = log(x+y)`, then:

Options

A. x+y=0
B. x=y
C. x=2, y=0
D. x =log y

Answer

To solve the equation \( \frac{1}{2} \log x + \frac{1}{2} \log y + \log 2 = \log(x+y) \), we can follow these steps:

### Step 1: Combine the logarithmic terms on the left side
Using the property of logarithms that states \( a \log b = \log(b^a) \), we can rewrite the left side:
\[ \frac{1}{2} \log x + \frac{1}{2} \log y = \log(x^{1/2}) + \log(y^{1/2}) = \log(\sqrt{x} \cdot \sqrt{y}) = \log(\sqrt{xy}) \]
Thus, the equation becomes:
\[ \log(\sqrt{xy}) + \log 2 = \log(x+y) \]

### Step 2: Combine the logarithms on the left side
Using the property \( \log a + \log b = \log(ab) \):
\[ \log(2\sqrt{xy}) = \log(x+y) \]

### Step 3: Remove the logarithm by exponentiating both sides
Since the logarithm function is one-to-one, we can equate the arguments:
\[ 2\sqrt{xy} = x + y \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (2\sqrt{xy})^2 = (x + y)^2 \]
This simplifies to:
\[ 4xy = x^2 + 2xy + y^2 \]

### Step 5: Rearrange the equation
Rearranging the equation leads to:
\[ 4xy - 2xy - x^2 - y^2 = 0 \]
This simplifies to:
\[ 2xy - x^2 - y^2 = 0 \]

### Step 6: Rearranging further
Rearranging gives:
\[ x^2 + y^2 - 2xy = 0 \]

### Step 7: Factor the equation
This can be factored as:
\[ (x - y)^2 = 0 \]

### Step 8: Solve for x and y
Taking the square root of both sides gives:
\[ x - y = 0 \quad \Rightarrow \quad x = y \]

### Conclusion
Thus, the solution to the equation is \( x = y \).

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