STATEMENT -1 : If f(x) `= log_(x^(2)) (log x) , " then" f'( e) = 1/e `
STATEMENT -2 : If a gt 0 , b gt 0 and ` a ne b` then ` log_(a) b = (log b)/(log a)`

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 expanation for statement-1
C. Statement-1 is True, Statement-2 is false
D. Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
We have the function:
\[ f(x) = \log_{x^2}(\log x) \]

Using the change of base formula for logarithms, we can rewrite this:
\[ f(x) = \frac{\log(\log x)}{\log(x^2)} \]
Since \(\log(x^2) = 2\log x\), we can simplify this further:
\[ f(x) = \frac{\log(\log x)}{2 \log x} \]

### Step 2: Differentiate \( f(x) \)
To find \( f'(x) \), we will use the quotient rule:
If \( f(x) = \frac{u}{v} \), then:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Here, let:
- \( u = \log(\log x) \)
- \( v = 2 \log x \)

Now, we need to find \( u' \) and \( v' \):
1. Differentiate \( u \):
\[ u' = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \]

2. Differentiate \( v \):
\[ v' = 2 \cdot \frac{1}{x} = \frac{2}{x} \]

Now applying the quotient rule:
\[ f'(x) = \frac{\left(\frac{1}{x \log x}\right)(2 \log x) - \left(\log(\log x)\right)\left(\frac{2}{x}\right)}{(2 \log x)^2} \]

This simplifies to:
\[ f'(x) = \frac{\frac{2}{x} - \frac{2 \log(\log x)}{x}}{4 (\log x)^2} \]
\[ = \frac{2(1 - \log(\log x))}{4x (\log x)^2} = \frac{1 - \log(\log x)}{2x (\log x)^2} \]

### Step 3: Evaluate \( f'(e) \)
Now we substitute \( x = e \):
\[ f'(e) = \frac{1 - \log(\log e)}{2e (\log e)^2} \]
Since \( \log e = 1 \), we have:
\[ \log(\log e) = \log(1) = 0 \]
Thus:
\[ f'(e) = \frac{1 - 0}{2e (1)^2} = \frac{1}{2e} \]

### Conclusion for Statement 1
The statement claims \( f'(e) = \frac{1}{e} \), but we found \( f'(e) = \frac{1}{2e} \). Therefore, Statement 1 is **false**.

### Step 4: Analyze Statement 2
The second statement is:
"If \( a > 0 \), \( b > 0 \) and \( a \neq b \), then \( \log_a b = \frac{\log b}{\log a} \)."

This is a well-known property of logarithms and is **true**.

### Final Result
- Statement 1: False
- Statement 2: True

### Summary
The correct answer is that Statement 1 is false and Statement 2 is true.