If 3 `log_(10) (x^(2) y) = 4 + 2 log_(10) x - log_(10) y`, where x and y are both + ve, and x - y = `2 sqrt(6)`, then the value of x is

Options

A. `4 + sqrt(6)`
B. `4 - sqrt(6)`
C. `sqrt(6)`
D. none of these

Answer

To solve the equation \( 3 \log_{10} (x^2 y) = 4 + 2 \log_{10} x - \log_{10} y \) with the conditions that \( x \) and \( y \) are both positive and \( x - y = 2 \sqrt{6} \), we will follow these steps:

### Step-by-Step Solution:

1. **Rewrite the Equation**:
Start with the given equation:
\[ 3 \log_{10} (x^2 y) = 4 + 2 \log_{10} x - \log_{10} y \]

2. **Use Logarithmic Properties**:
Apply the properties of logarithms:
\[ 3 \log_{10} (x^2 y) = 3 (\log_{10} x^2 + \log_{10} y) = 3 (2 \log_{10} x + \log_{10} y) = 6 \log_{10} x + 3 \log_{10} y \]
Therefore, we can rewrite the left side:
\[ 6 \log_{10} x + 3 \log_{10} y = 4 + 2 \log_{10} x - \log_{10} y \]

3. **Rearranging the Equation**:
Move all terms to one side:
\[ 6 \log_{10} x + 3 \log_{10} y - 2 \log_{10} x + \log_{10} y - 4 = 0 \]
Combine like terms:
\[ (6 - 2) \log_{10} x + (3 + 1) \log_{10} y - 4 = 0 \]
This simplifies to:
\[ 4 \log_{10} x + 4 \log_{10} y - 4 = 0 \]

4. **Factor Out the Common Term**:
Factor out 4:
\[ 4 (\log_{10} x + \log_{10} y - 1) = 0 \]
Thus, we have:
\[ \log_{10} x + \log_{10} y = 1 \]

5. **Convert to Exponential Form**:
This implies:
\[ \log_{10} (xy) = 1 \implies xy = 10 \]

6. **Use the Given Condition**:
We also have the condition \( x - y = 2 \sqrt{6} \).

7. **Express y in terms of x**:
From \( xy = 10 \), we can express \( y \):
\[ y = \frac{10}{x} \]

8. **Substitute into the Condition**:
Substitute \( y \) into the condition \( x - y = 2 \sqrt{6} \):
\[ x - \frac{10}{x} = 2 \sqrt{6} \]

9. **Multiply through by x**:
Multiply through by \( x \) to eliminate the fraction:
\[ x^2 - 10 = 2 \sqrt{6} x \]

10. **Rearrange into Standard Form**:
Rearranging gives:
\[ x^2 - 2 \sqrt{6} x - 10 = 0 \]

11. **Use the Quadratic Formula**:
Apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2\sqrt{6}, c = -10 \):
\[ x = \frac{2\sqrt{6} \pm \sqrt{(-2\sqrt{6})^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \]
\[ = \frac{2\sqrt{6} \pm \sqrt{24 + 40}}{2} \]
\[ = \frac{2\sqrt{6} \pm \sqrt{64}}{2} \]
\[ = \frac{2\sqrt{6} \pm 8}{2} \]
\[ = \sqrt{6} \pm 4 \]

12. **Determine the Positive Solution**:
Since \( x \) must be positive:
\[ x = \sqrt{6} + 4 \]
The negative solution \( \sqrt{6} - 4 \) is not valid since \( \sqrt{6} \approx 2.45 \).

### Final Answer:
Thus, the value of \( x \) is:
\[ \boxed{\sqrt{6} + 4} \]