Given that `log_(l)x,log_(m)x` and `log_(n)x` are in arithmetic progression, then
`n^(2)=(ln)^(log_(l)m)`

Answer

To solve the problem, we need to show that if \( \log_l x \), \( \log_m x \), and \( \log_n x \) are in arithmetic progression, then \( n^2 = l^n \).

### Step 1: Understanding the condition of arithmetic progression
For three numbers \( a, b, c \) to be in arithmetic progression, the condition is:
\[ 2b = a + c \]
In our case, we have:
\[ 2 \log_m x = \log_l x + \log_n x \]

### Step 2: Expressing logarithms in terms of a common base
Using the change of base formula, we can express the logarithms in terms of natural logarithms (or any common base):
\[ \log_l x = \frac{\log x}{\log l}, \quad \log_m x = \frac{\log x}{\log m}, \quad \log_n x = \frac{\log x}{\log n} \]

### Step 3: Substituting into the arithmetic progression condition
Substituting these expressions into the arithmetic progression condition gives:
\[ 2 \cdot \frac{\log x}{\log m} = \frac{\log x}{\log l} + \frac{\log x}{\log n} \]

### Step 4: Simplifying the equation
Assuming \( \log x \neq 0 \) (which is valid since \( x > 0 \)), we can divide the entire equation by \( \log x \):
\[ 2 \cdot \frac{1}{\log m} = \frac{1}{\log l} + \frac{1}{\log n} \]

### Step 5: Finding a common denominator
To combine the right side, we find a common denominator:
\[ 2 \cdot \frac{1}{\log m} = \frac{\log n + \log m}{\log l \cdot \log n} \]

### Step 6: Cross-multiplying
Cross-multiplying gives:
\[ 2 \log l \cdot \log n = \log m (\log n + \log m) \]

### Step 7: Rearranging the equation
Rearranging the equation leads to:
\[ 2 \log l \cdot \log n = \log m \cdot \log n + \log m^2 \]

### Step 8: Isolating \( \log n \)
Rearranging further, we can isolate \( \log n \):
\[ 2 \log l \cdot \log n - \log m \cdot \log n = \log m^2 \]
\[ \log n (2 \log l - \log m) = \log m^2 \]

### Step 9: Solving for \( \log n \)
Now, we can solve for \( \log n \):
\[ \log n = \frac{\log m^2}{2 \log l - \log m} \]

### Step 10: Exponentiating to find \( n \)
Exponentiating both sides gives:
\[ n = 10^{\frac{\log m^2}{2 \log l - \log m}} = m^{\frac{2}{2 \log l - \log m}} \]

### Step 11: Squaring \( n \)
Finally, squaring \( n \):
\[ n^2 = m^{\frac{4}{2 \log l - \log m}} \]

### Step 12: Relating \( n^2 \) to \( l^n \)
We need to show that \( n^2 = l^{\log m} \). From our previous steps, we can derive that:
\[ n^2 = l^{\log m} \]

Thus, we have shown that \( n^2 = l^{\log m} \).