If `a > 1, b > 1,` then the minimum value of `log_b a + log_a b` is

Options

A. 0
B. 1
C. 2
D. 4

Answer

To find the minimum value of \( \log_b a + \log_a b \) given that \( a > 1 \) and \( b > 1 \), we can follow these steps:

### Step 1: Rewrite the logarithmic expressions
We can use the change of base formula for logarithms, which states that:
\[ \log_b a = \frac{\log a}{\log b} \quad \text{and} \quad \log_a b = \frac{\log b}{\log a} \]
Thus, we can rewrite the expression:
\[ \log_b a + \log_a b = \frac{\log a}{\log b} + \frac{\log b}{\log a} \]

### Step 2: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
According to the AM-GM inequality, for any positive numbers \( x \) and \( y \):
\[ \frac{x + y}{2} \geq \sqrt{xy} \]
Let \( x = \frac{\log a}{\log b} \) and \( y = \frac{\log b}{\log a} \). Then we have:
\[ \frac{\frac{\log a}{\log b} + \frac{\log b}{\log a}}{2} \geq \sqrt{\frac{\log a}{\log b} \cdot \frac{\log b}{\log a}} \]
The right-hand side simplifies to:
\[ \sqrt{1} = 1 \]
Thus, we can write:
\[ \frac{\log_b a + \log_a b}{2} \geq 1 \]

### Step 3: Multiply by 2
Multiplying both sides of the inequality by 2 gives:
\[ \log_b a + \log_a b \geq 2 \]

### Step 4: Conclusion
The minimum value of \( \log_b a + \log_a b \) is therefore:
\[ \boxed{2} \]