Statement-1: `"log"_(10)x lt "log"_(pi) x lt "log"_(e) x lt "log"_(2) x`
Statement-2: `x lt y rArr "log"_(a) x gt "log"_(a) y " when " 0 lt a lt 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 given statements, we will analyze each statement step by step.

### Statement 1:
We need to verify the inequality:
\[ \log_{10} x < \log_{\pi} x < \log_{e} x < \log_{2} x \]

#### Step 1: Rewrite the logarithms using the change of base formula.
Using the change of base formula, we can express the logarithms as:
\[ \log_{b} x = \frac{\log_{10} x}{\log_{10} b} \]
Thus, we can rewrite the inequality as:
\[ \frac{\log_{10} x}{\log_{10} 10} < \frac{\log_{10} x}{\log_{10} \pi} < \frac{\log_{10} x}{\log_{10} e} < \frac{\log_{10} x}{\log_{10} 2} \]

#### Step 2: Simplify the inequality.
Since \(\log_{10} 10 = 1\), we can simplify the inequality to:
\[ \log_{10} x < \frac{\log_{10} x}{\log_{10} \pi} < \frac{\log_{10} x}{\log_{10} e} < \frac{\log_{10} x}{\log_{10} 2} \]

#### Step 3: Analyze the denominators.
The values of \(\log_{10} 10\), \(\log_{10} \pi\), \(\log_{10} e\), and \(\log_{10} 2\) are:
- \(\log_{10} 10 = 1\)
- \(\log_{10} \pi \approx 0.497\)
- \(\log_{10} e \approx 0.434\)
- \(\log_{10} 2 \approx 0.301\)

Since \(1 > \pi > e > 2\), we can conclude that:
\[ \log_{10} 10 < \log_{10} \pi < \log_{10} e < \log_{10} 2 \]

#### Step 4: Reverse the inequality.
Since the logarithm of \(x\) is the same in the numerators, the inequality holds true in reverse order:
\[ \log_{10} x < \log_{\pi} x < \log_{e} x < \log_{2} x \]
This confirms that Statement 1 is true.

### Statement 2:
We need to verify the statement:
\[ x < y \Rightarrow \log_{a} x > \log_{a} y \text{ when } 0 < a < 1 \]

#### Step 5: Analyze the logarithmic property.
For a base \(a\) where \(0 < a < 1\), the logarithmic function is decreasing. Therefore, if \(x < y\), then:
\[ \log_{a} x > \log_{a} y \]
This confirms that Statement 2 is also true.

### Conclusion:
Both statements are true.