Statement-1: The solution set of the equation `"log"_(x) 2 xx "log"_(2x) 2 = "log"_(4x) 2 "is" {2^(-sqrt(2)), 2^(sqrt(2))}.` Statement-2 : `"log"_(b)a = (1)/("log"_(a)b) " and log"_(a) xy = "log"_(a) x + "log"_(a)y`

Options

A. Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is true, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem step by step, we will analyze the given equation and verify the statements.

### Step 1: Rewrite the given equation
We start with the equation:
\[ \log_{x} 2 \cdot \log_{2x} 2 = \log_{4x} 2 \]

### Step 2: Apply the change of base formula
Using the change of base formula, we can rewrite the logarithms:
\[ \log_{x} 2 = \frac{\log 2}{\log x}, \quad \log_{2x} 2 = \frac{\log 2}{\log(2x)}, \quad \log_{4x} 2 = \frac{\log 2}{\log(4x)} \]
Substituting these into the equation gives:
\[ \frac{\log 2}{\log x} \cdot \frac{\log 2}{\log(2) + \log(x)} = \frac{\log 2}{\log(4) + \log(x)} \]

### Step 3: Simplify the equation
This simplifies to:
\[ \frac{(\log 2)^2}{\log x (\log 2 + \log x)} = \frac{\log 2}{2 \log 2 + \log x} \]

### Step 4: Cross-multiply to eliminate fractions
Cross-multiplying gives:
\[ (\log 2)^2 (2 \log 2 + \log x) = \log 2 \cdot \log x (\log 2 + \log x) \]

### Step 5: Expand both sides
Expanding both sides results in:
\[ 2(\log 2)^3 + (\log 2)^2 \log x = \log 2 \cdot \log x \cdot \log 2 + \log 2 \cdot \log x \cdot \log x \]
This simplifies to:
\[ 2(\log 2)^3 + (\log 2)^2 \log x = (\log 2)^2 \log x + (\log x)^2 \log 2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 2(\log 2)^3 = (\log x)^2 \log 2 \]

### Step 7: Divide by \(\log 2\) (assuming \(\log 2 \neq 0\))
Dividing both sides by \(\log 2\):
\[ 2(\log 2)^2 = (\log x)^2 \]

### Step 8: Taking square roots
Taking the square root of both sides yields:
\[ \log x = \pm \sqrt{2} \log 2 \]

### Step 9: Exponentiating to solve for \(x\)
Exponentiating gives:
\[ x = 2^{\sqrt{2}} \quad \text{or} \quad x = 2^{-\sqrt{2}} \]

### Conclusion
Thus, the solution set is:
\[ \{2^{-\sqrt{2}}, 2^{\sqrt{2}}\} \]
This confirms that Statement-1 is true.

### Verification of Statement-2
Statement-2 states:
1. \(\log_{b} a = \frac{1}{\log_{a} b}\) - This is true.
2. \(\log_{a} (xy) = \log_{a} x + \log_{a} y\) - This is also true.

Both statements are valid properties of logarithms.

### Final Answer
Both statements are true, and Statement-1 is verified as correct.