Statement-1: `3^("log"_(2) 7) -7^("log"_(2)3) = 0`
Statement-2: `x^("log"_(a)y) = y^("log"_(a)x), " where "x gt 0, y gt 0" and "a gt 0, a ne 1`

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, we need to analyze both statements step by step.

### Step 1: Analyze Statement-1
We are given the expression:
\[ 3^{\log_2 7} - 7^{\log_2 3} = 0 \]

This can be rewritten as:
\[ 3^{\log_2 7} = 7^{\log_2 3} \]

### Step 2: Apply the Logarithmic Property
We can use the property of logarithms which states:
\[ x^{\log_a y} = y^{\log_a x} \]
for \( x > 0, y > 0 \), and \( a > 0 \) (where \( a \neq 1 \)).

In our case, we can let \( x = 3 \), \( y = 7 \), and \( a = 2 \):
- Thus, we have:
\[ 3^{\log_2 7} = 7^{\log_2 3} \]

### Step 3: Conclusion for Statement-1
Since we have shown that:
\[ 3^{\log_2 7} = 7^{\log_2 3} \]
This implies that:
\[ 3^{\log_2 7} - 7^{\log_2 3} = 0 \]
Therefore, Statement-1 is **true**.

### Step 4: Analyze Statement-2
Statement-2 is:
\[ x^{\log_a y} = y^{\log_a x} \]
where \( x > 0, y > 0 \), and \( a > 0 \) (with \( a \neq 1 \)).

This statement is a known property of logarithms and is indeed true.

### Step 5: Conclusion for Statement-2
Since we have validated that the property holds true for the given conditions, Statement-2 is also **true**.

### Final Conclusion
Both Statement-1 and Statement-2 are true, and Statement-2 was used to prove Statement-1.

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