If a and b are positive real numbers other than unity, then the least value of `|"log"_(b) a + "log"_(a) b|`, is

Options

A. 0
B. 1
C. 2
D. none of these

Explanation

To find the least value of \( | \log_b a + \log_a b | \), we can follow these steps:

### Step 1: Rewrite the expression using logarithmic properties
We know that:
\[ \log_a b = \frac{1}{\log_b a} \]
Thus, we can rewrite the expression:
\[ \log_b a + \log_a b = \log_b a + \frac{1}{\log_b a} \]

### Step 2: Let \( x = \log_b a \)
Now, we can substitute \( x \) into the expression:
\[ x + \frac{1}{x} \]

### Step 3: Analyze the function \( f(x) = x + \frac{1}{x} \)
We need to find the minimum value of \( |f(x)| \). Since \( a \) and \( b \) are positive real numbers other than unity, \( x \) can take any positive value.

### Step 4: Find the minimum value of \( f(x) \)
To find the minimum value of \( f(x) \), we can use calculus. We differentiate \( f(x) \):
\[ f'(x) = 1 - \frac{1}{x^2} \]
Setting the derivative to zero to find critical points:
\[ 1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = 1 \quad (\text{since } x > 0) \]

### Step 5: Evaluate \( f(x) \) at the critical point
Now, we evaluate \( f(1) \):
\[ f(1) = 1 + \frac{1}{1} = 2 \]

### Step 6: Determine the minimum value of \( |f(x)| \)
Since \( f(x) \) is always greater than or equal to 2 for \( x > 0 \), we have:
\[ |f(x)| \geq 2 \]

### Conclusion
Thus, the least value of \( | \log_b a + \log_a b | \) is:
\[ \boxed{2} \]