Statement-1(Assertion) and Statement-2 (reason) Each of these question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1
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
Statement -1 `log_10xltlog_3xltlog_exltlog_2x` `(xgt0,xne1)`
Statment If `0ltxlt1`, then `log_xagtlog_xbimplies0ltaltb`.

Answer

To solve the problem, we need to analyze both statements and determine their truth values as well as the relationship between them.

### Step 1: Analyze Statement 1
Statement 1: \( \log_{10} x < \log_{3} x < \log_{e} x < \log_{2} x \) for \( x > 0 \) and \( x \neq 1 \).

1. **Understanding the logarithmic properties**:
- We know that for \( x > 1 \), logarithmic functions are increasing, and for \( 0 < x < 1 \), they are decreasing.
- The bases of the logarithms are \( 10, 3, e, 2 \) where \( 2 < e < 3 < 10 \).

2. **Reciprocal of logarithms**:
- Taking the reciprocal of the logarithms changes the inequality direction:
\[ \frac{1}{\log_{2}} > \frac{1}{\log_{e}} > \frac{1}{\log_{3}} > \frac{1}{\log_{10}} \]

3. **Multiplying by \( \log x \)**:
- If \( \log x > 0 \) (i.e., \( x > 1 \)), the inequality remains the same:
\[ \log_{2} x > \log_{e} x > \log_{3} x > \log_{10} x \]
- If \( \log x < 0 \) (i.e., \( 0 < x < 1 \)), the inequality reverses:
\[ \log_{2} x < \log_{e} x < \log_{3} x < \log_{10} x \]

4. **Conclusion for Statement 1**:
- The statement \( \log_{10} x < \log_{3} x < \log_{e} x < \log_{2} x \) is true for \( x > 1 \) and false for \( 0 < x < 1 \).

### Step 2: Analyze Statement 2
Statement 2: If \( 0 < x < 1 \), then \( \log_{a} x > \log_{b} x \) implies \( 0 < a < b \).

1. **Understanding the implication**:
- If \( \log_{a} x > \log_{b} x \), we can rewrite it as:
\[ \frac{\log a}{\log x} > \frac{\log b}{\log x} \]
- Since \( \log x < 0 \) for \( 0 < x < 1 \), multiplying by \( \log x \) reverses the inequality:
\[ \log a < \log b \]

2. **Taking antilogarithm**:
- Taking the antilogarithm gives us \( a < b \), which implies \( 0 < a < b \) since logarithms are defined for positive numbers.

3. **Conclusion for Statement 2**:
- Statement 2 is true.

### Final Conclusion
- **Statement 1** is false for \( 0 < x < 1 \).
- **Statement 2** is true.

### Answer Choice
- The correct choice is: **(c) Statement 1 is true, Statement 2 is false.**