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

Options

A. x+y = 0
B. x-y =0
C. xy =1
D. `x^(2) + xy + y^(2) =0`

Answer

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

### Step 1: Combine the logarithmic terms on the left side
Using the property of logarithms that states \( n \log a = \log(a^n) \), we can rewrite the left-hand side:

\[ \frac{1}{2} \log x = \log(x^{1/2}) = \log(\sqrt{x}) \]
\[ \frac{1}{2} \log y = \log(y^{1/2}) = \log(\sqrt{y}) \]

Thus, we can rewrite the left-hand side as:

\[ \log(\sqrt{x}) + \log(\sqrt{y}) + \log(2) \]

### Step 2: Use the property of logarithms to combine
Now we can use the property \( \log a + \log b = \log(ab) \):

\[ \log(\sqrt{x} \cdot \sqrt{y} \cdot 2) = \log(2\sqrt{xy}) \]

So the equation becomes:

\[ \log(2\sqrt{xy}) = \log(x+y) \]

### Step 3: Remove the logarithms
Since the logarithms are equal, we can set the arguments equal to each other:

\[ 2\sqrt{xy} = x + y \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:

\[ (2\sqrt{xy})^2 = (x + y)^2 \]

This simplifies to:

\[ 4xy = x^2 + 2xy + y^2 \]

### Step 5: Rearrange the equation
Now, we can rearrange the equation:

\[ 4xy - 2xy = x^2 + y^2 \]

This simplifies to:

\[ 2xy = x^2 + y^2 \]

### Step 6: Rearranging further
We can rewrite this as:

\[ x^2 - 2xy + y^2 = 0 \]

### 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 us:

\[ x - y = 0 \implies x = y \]

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

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