If `log_(l)p, log_(m)p and log_(n)p` are in AP, then `(ln)^(log_(l)m)` = ______.

Options

A. `n^(2)`
B. `m^(2)`
C. `l^(2)`
D. `p^(2)`.

Explanation

To solve the problem, we need to show that if \( \log_{l} p \), \( \log_{m} p \), and \( \log_{n} p \) are in arithmetic progression (AP), then \( (ln)^{\log_{l} m} = ? \).

### Step-by-step Solution:

1. **Understanding the Arithmetic Progression**:
Since \( \log_{l} p \), \( \log_{m} p \), and \( \log_{n} p \) are in AP, we can express this condition mathematically:
\[ 2 \log_{m} p = \log_{l} p + \log_{n} p \]

2. **Using the Change of Base Formula**:
We can rewrite the logarithms using the change of base formula:
\[ \log_{l} p = \frac{\log p}{\log l}, \quad \log_{m} p = \frac{\log p}{\log m}, \quad \log_{n} p = \frac{\log p}{\log n} \]

3. **Substituting into the AP Condition**:
Substitute these expressions into the AP condition:
\[ 2 \left(\frac{\log p}{\log m}\right) = \frac{\log p}{\log l} + \frac{\log p}{\log n} \]

4. **Eliminating \(\log p\)**:
Assuming \( \log p \neq 0 \), we can divide both sides by \( \log p \):
\[ 2 \cdot \frac{1}{\log m} = \frac{1}{\log l} + \frac{1}{\log n} \]

5. **Finding a Common Denominator**:
The common denominator for the right-hand side is \( \log l \cdot \log n \):
\[ 2 \cdot \frac{\log l \cdot \log n}{\log m \cdot \log l \cdot \log n} = \frac{\log n + \log l}{\log l \cdot \log n} \]

6. **Cross Multiplying**:
Cross-multiplying gives:
\[ 2 \log l \cdot \log n = \log n + \log l \]

7. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ 2 \log l \cdot \log n - \log n - \log l = 0 \]
This can be factored as:
\[ (\log l - 1)(\log n - 1) = 0 \]

8. **Solving for \( \log l \) and \( \log n \)**:
This implies either \( \log l = 1 \) or \( \log n = 1 \). Thus, we can conclude that:
\[ l = e \quad \text{or} \quad n = e \]

9. **Finding \( (ln)^{\log_{l} m} \)**:
Now, we need to compute \( (ln)^{\log_{l} m} \):
\[ (ln)^{\log_{l} m} = (e \cdot n)^{\log_{l} m} \]
Since \( l = e \):
\[ (en)^{\log_{e} m} = (en)^{\frac{\log m}{\log e}} = (en)^{\log m} \]

10. **Final Expression**:
Therefore, we can simplify this to:
\[ n^{\log m} \cdot e^{\log m} = n^{\log m} \cdot m \]

### Final Answer:
Thus, the final expression is:
\[ (ln)^{\log_{l} m} = n^{\log m} \]