How many ordered pair (s) satisfy `log (x ^(3) + (1)/(3) y ^(3) + (1)/(9))= log x + log y`

Answer

To solve the equation \( \log(x^3 + \frac{1}{3}y^3 + \frac{1}{9}) = \log x + \log y \), we can 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-hand side of the equation:
\[ \log(x^3 + \frac{1}{3}y^3 + \frac{1}{9}) = \log(xy) \]

### Step 2: Remove the logarithm
Since the logarithm function is one-to-one, we can equate the arguments:
\[ x^3 + \frac{1}{3}y^3 + \frac{1}{9} = xy \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ x^3 + \frac{1}{3}y^3 + \frac{1}{9} - xy = 0 \]

### Step 4: Recognize the form of the equation
This can be recognized as a form of the identity for cubes:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]
where \( a = x \), \( b = \frac{1}{3}y \), and \( c = \frac{1}{9} \).

### Step 5: Apply the identity
From the identity, we know that either:
1. \( a + b + c = 0 \) or
2. \( a = b = c \).

### Step 6: Solve for \( a + b + c = 0 \)
Substituting our values:
\[ x + \frac{1}{3}y + \frac{1}{9} = 0 \]
This implies:
\[ x + \frac{1}{3}y = -\frac{1}{9} \]
Since \( x \) and \( y \) must be positive, this case does not yield valid solutions.

### Step 7: Solve for \( a = b = c \)
Setting \( a = b = c \):
\[ x = \frac{1}{3}y = \frac{1}{9} \]
From \( x = \frac{1}{9} \), we find \( y \):
\[ \frac{1}{3}y = \frac{1}{9} \implies y = \frac{1}{3} \]

### Step 8: Identify the ordered pair
Thus, we have the ordered pair:
\[ (x, y) = \left(\frac{1}{9}, \frac{1}{3}\right) \]

### Step 9: Check for validity
We need to ensure that both \( x \) and \( y \) are positive and that they satisfy the original logarithmic equation. Since both values are positive, we confirm that:
\[ \log\left(\frac{1}{9}^3 + \frac{1}{3}\left(\frac{1}{3}\right)^3 + \frac{1}{9}\right) = \log\left(\frac{1}{9}\right) + \log\left(\frac{1}{3}\right) \]
is valid.

### Conclusion
Thus, the only ordered pair that satisfies the equation is:
\[ \left(\frac{1}{9}, \frac{1}{3}\right) \]
So, the answer is that there is **1 ordered pair**.

---